Reference - What does this error mean in PHP?

Reference - What does this error mean in PHP?

This is a number of answers about warnings, errors and notices you might encounter while programming PHP and have no clue how to fix. This is also a Community Wiki, so everyone is invited to participate in adding to and maintaining this list.

Questions like "Headers already sent" or "Calling a member of a non-object" pop up frequently on Stack Overflow. The root cause of those questions is always the same. So the answers to those questions typically repeat them and then show the OP which line to change in his/her particular case. These answers do not add any value to the site because they only apply to the OP's particular code. Other users having the same error can not easily read the solution out of it because they are too localized. That is sad, because once you understood the root cause, fixing the error is trivial. Hence, this list tries to explain the solution in a general way to apply.

If your question has been marked as a duplicate of this, please find your error message below and apply the fix to your code. The answers usually contain further links to investigate in case it shouldn't be clear from the general answer alone.

If you want to contribute, please add your "favorite" error message, warning or notice, one per answer, a short description what it means (even if it is only highlighting terms to their manual page), a possible solution or debugging approach and a listing of existing Q&A that are of value. Also, feel free to improve any existing answers.

Also see

@PeterSmit: The php-errors tag has no wiki summary, can you help us create it?
– hakre
Oct 8 '12 at 12:44

Also, to move discussion out of comments, please go to this meta question
– Earlz
Oct 8 '12 at 13:56

X-Ref: PHP Parse/Syntax Errors; and How to solve them?; Elsewhere: Common PHP Errors and Solutions
– hakre
Aug 11 '13 at 21:47

see this coding.smashingmagazine.com/2011/11/30/… and this also mediawiki.org/wiki/Manual:Errors_and_symptoms
– krishna
Aug 27 '13 at 14:26

X-Ref: Fixing PHP Errors (May 2013; by Jason McCreary)
– hakre
Sep 3 '13 at 16:37

30 Answers
30

Happens when your script tries to send a HTTP header to the client but there already was output before, which resulted in headers to be already sent to the client.

This is an E_WARNING and it will not stop the script.

E_WARNING

A typical example would be a template file like this:

<html> <?php session_start(); ?> <head><title>My Page</title> </html> ...

The session_start() function will try to send headers with the session cookie to the client. But PHP already sent headers when it wrote the <html> element to the output stream. You'd have to move the session_start() to the top.

session_start()

<html>

session_start()

You can solve this by going through the lines before the code triggering the Warning and check where it outputs. Move any header sending code before that code.

An often overlooked output is new lines after PHP's closing ?>. It is considered a standard practice to omit ?> when it is the last thing in the file. Likewise, another common cause for this warning is when the opening <?php has an empty space, line, or invisible character before it, causing the webserver to send the headers and the whitespace/newline thus when PHP starts parsing won't be able to submit any header.

?>

<?php

If your file has more than one <?php ... ?> code block in it, you should not have any spaces in between them. (Note: You might have multiple blocks if you had code that was automatically constructed)

<?php ... ?>

Also make sure you don't have any Byte Order Marks in your code, for example when the encoding of the script is UTF-8 with BOM.

Related Questions:

If you are using WordPress, check the theme files. When I upgraded a site to a new version of WordPress, I was unable to update the theme because it has not been updated in several years. This problem cropped up. It turned out that the functions.php file had more than one <? ?> block with spaces in between.
– Roy Leban
Mar 31 '13 at 8:41

@RoyLeban "If your file has more than one block in it..." I'm not sure what this means. What is a "block"? Would one block consist of <?php ?> and so "more than one block" would be <?php ?> <?php ?>?
– Andrew Fox
Jan 30 '14 at 8:40

<?php ?>

<?php ?> <?php ?>

Please turn on 'output buffering' feature in PHP.ini configuration file if possible.It is used for sloving this issue.It sends the html file is saved in the output buffer and sent to the client only after the script stops, so if two headers are issued at different location then the old header will be replaced new header.
– Nidhin David
May 13 '14 at 17:15

Happens with code similar to xyz->method() where xyz is not an object and therefore that method can not be called.

xyz->method()

xyz

method

This is a fatal error which will stop the script (forward compatibility notice: It will become a catchable error starting with PHP 7).

Most often this is a sign that the code has missing checks for error conditions. Validate that an object is actually an object before calling its methods.

A typical example would be

// ... some code using PDO $statement = $pdo->prepare('invalid query', ...); $statement->execute(...);

In the example above, the query cannot be prepared and prepare() will assign false to $statement. Trying to call the execute() method will then result in the Fatal Error because false is a "non-object" because the value is a boolean.

prepare()

false

$statement

execute()

false

Figure out why your function returned a boolean instead of an object. For example, check the $pdo object for the last error that occurred. Details on how to debug this will depend on how errors are handled for the particular function/object/class in question.

$pdo

If even the ->prepare is failing then your $pdo database handle object didn't get passed into the current scope. Find where it got defined. Then pass it as a parameter, store it as property, or share it via the global scope.

->prepare

$pdo

Another problem may be conditionally creating an object and then trying to call a method outside that conditional block. For example

if ($someCondition) { $myObj = new MyObj(); } // ... $myObj->someMethod();

By attempting to execute the method outside the conditional block, your object may not be defined.

Related Questions:

Also known as the White Page Of Death or White Screen Of Death. This happens when error reporting is turned off and a fatal error (often syntax error) occurred.

If you have error logging enabled, you will find the concrete error message in your error log. This will usually be in a file called "php_errors.log", either in a central location (e.g. /var/log/apache2 on many Linux environments) or in the directory of the script itself (sometimes used in a shared hosting environment).

/var/log/apache2

Sometimes it might be more straightforward to temporarily enable the display of errors. The white page will then display the error message. Take care because these errors are visible to everybody visiting the website.

This can be easily done by adding at the top of the script the following PHP code:

ini_set('display_errors', 1); error_reporting(~0);

The code will turn on the display of errors and set reporting to the highest level.

Since the ini_set() is executed at runtime it has no effects on parsing/syntax errors. Those errors will appear in the log. If you want to display them in the output as well (e.g. in a browser) you have to set the display_startup_errors directive to true. Do this either in the php.ini or in a .htaccess or by any other method that affects the configuration before runtime.

ini_set()

display_startup_errors

true

php.ini

.htaccess

You can use the same methods to set the log_errors and error_log directives to choose your own log file location.

Looking in the log or using the display, you will get a much better error message and the line of code where your script comes to halt.

Related Questions:

First and foremost:

Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.

mysql_*

This happens when you try to fetch data from the result of mysql_query but the query failed.

mysql_query

This is a warning and won't stop the script, but will make your program wrong.

You need to check the result returned by mysql_query by

mysql_query

$res = mysql_query($sql); if (!$res) { die(mysql_error()); } // after checking, do the fetch

Related Questions:

Related Errors:

Other mysql* functions that also expect a mysql result resource as parameter will produce the same error for the same reason.

mysql*

Just a note. If mysql_query isn't bad enough, adding or die on top of it is adding insult to injury.
– Madara Uchiha♦
Oct 7 '12 at 22:16

mysql_query

or die

The problem I encounter is $res = mysql_query($query) returns 1 if query is successful so it is considered true. Therefore when passing the result of mysql_query to mysql_fetch_array() the notice shows.
– mboy
Nov 27 '16 at 16:09

$res = mysql_query($query)

mysql_query

mysql_fetch_array()

@mboy For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error. For other type of SQL statements, INSERT, UPDATE, DELETE, DROP, etc, mysql_query() returns TRUE on success or FALSE on error.
– xdazz
Nov 28 '16 at 0:47

@xdazz That is the problem I am facing, Insert update returns TRUE so I cannot get rid of this error mysql_fetch_array() expects parameter 1 to be resource, boolean given in select please see - gist.github.com/romelemperado/93af4cdbd44ebf3a07cbfa0e3fc539d7 Any suggestion to git rid of this error?
– mboy
Nov 28 '16 at 4:26

mysql_fetch_array() expects parameter 1 to be resource, boolean given in select

@mboy mysql_fetch_array() is for select query, for insert and update, you don't need to fetch the result set (and there is no result set let you fetch).
– xdazz
Nov 28 '16 at 4:35

mysql_fetch_array()

$this is a special variable in PHP which can not be assigned. If it is accessed in a context where it does not exist, this fatal error is given.

$this

This error can occur:

If a non-static method is called statically. Example:

class Foo { protected $var; public function __construct($var) { $this->var = $var; } public static function bar () { // ^^^^^^ echo $this->var; // ^^^^^ } } Foo::bar();

How to fix: review your code again, $this can only be used in object context, and should never be used in a static method. Also, a static method should not access the non-static property. Use self::$static_property to access the static property.

$this

self::$static_property

If code from a class method has been copied over into a normal function or just the global scope and keeping the $this special variable.
How to fix: Review the code and replace $this with a different substitution variable.

$this

Related Questions:

You might also want to mention how this works w/ closures (even in non-static methods) and how it's "fixed" in 5.4.
– Kendall Hopkins
Oct 7 '12 at 18:12

@hakre I was talking about a static call inside a Closure. Like $closure = function() { self::method(); }.
– Kendall Hopkins
Oct 9 '12 at 16:55

$closure = function() { self::method(); }

@KendallHopkins: That is a different error: "Fatal error: Cannot access self:: when no class scope is active" However with $this you can trigger the bespoken "Fatal error: Using $this when not in object context" : $closure = function() { $this->method(); };
– hakre
Oct 9 '12 at 17:02

$this

$closure = function() { $this->method(); };

Happens when you try to call a function that is not defined yet. Common causes include missing extensions and includes, conditional function declaration, function in function declaration or simple typos.

Example 1 - Conditional Function Declaration

$someCondition = false; if ($someCondition === true) { function fn() { return 1; } } echo fn(); // triggers error

In this case, fn() will never be declared because $someCondition is not true.

fn()

$someCondition

Example 2 - Function in Function Declaration

function createFn() { function fn() { return 1; } } echo fn(); // triggers error

In this case, fn will only be declared once createFn() gets called. Note that subsequent calls to createFn() will trigger an error about Redeclaration of an Existing function.

fn

createFn()

You may also see this for a PHP built-in function. Try searching for the function in the official manual, and check what "extension" (PHP module) it belongs to, and what versions of PHP support it.

In case of a missing extension, install that extension and enable it in php.ini. Refer to the Installation Instructions in the PHP Manual for the extension your function appears in. You may also be able to enable or install the extension using your package manager (e.g. apt in Debian or Ubuntu, yum in Red Hat or CentOS), or a control panel in a shared hosting environment.

apt

yum

If the function was introduced in a newer version of PHP from what you are using, you may find links to alternative implementations in the manual or its comment section. If it has been removed from PHP, look for information about why, as it may no longer be necessary.

In case of missing includes, make sure to include the file declaring the function before calling the function.

In case of typos, fix the typo.

Related Questions:

Happens when you have T_XXX token in unexpected place, unbalanced (superfluous) parentheses, use of short tag without activating it in php.ini, and many more.

T_XXX

Related Questions:

For further help see:

This usually happens when using a function directly with empty.

empty

Example:

if (empty(is_null(null))) { echo 'empty'; }

This is because empty is a language construct and not a function, it cannot be called with an expression as its argument in PHP versions before 5.5. Prior to PHP 5.5, the argument to empty() must be a variable, but an arbitrary expression (such as a the return value of a function) is permissible in PHP 5.5+.

empty

empty()

empty, despite its name, does not actually check if a variable is "empty". Instead, it checks if a variable doesn't exist, or == false. Expressions (like is_null(null) in the example) will always be deemed to exist, so here empty is only checking if it is equal to false. You could replace empty() here with !, e.g. if (!is_null(null)), or explicitly compare to false, e.g. if (is_null(null) == false).

empty

== false

is_null(null)

empty

empty()

!

if (!is_null(null))

if (is_null(null) == false)

Related Questions:

This error is often caused because you forgot to properly escape the data passed to a MySQL query.

An example of what not to do (the "Bad Idea"):

$query = "UPDATE `posts` SET my_text='{$_POST['text']}' WHERE id={$_GET['id']}"; mysqli_query($db, $query);

This code could be included in a page with a form to submit, with an URL such as http://example.com/edit.php?id=10 (to edit the post n°10)

What will happen if the submitted text contains single quotes? $query will end up with:

$query

$query = "UPDATE `posts` SET my_text='I'm a PHP newbie' WHERE id=10';

And when this query is sent to MySQL, it will complain that the syntax is wrong, because there is an extra single quote in the middle.

To avoid such errors, you MUST always escape the data before use in a query.

Escaping data before use in a SQL query is also very important because if you don't, your script will be open to SQL injections. An SQL injection may cause alteration, loss or modification of a record, a table or an entire database. This is a very serious security issue!

Documentation:

mysql_real_escape_string()

mysqli_real_escape_string()

In addition, if you don't your site will be hacked by bots automatically
– apscience
Oct 7 '12 at 22:32

@gladoscc Click "edit" and modify the answer. I am aware it can be improved.
– Jocelyn
Oct 8 '12 at 0:50

Or use a prepared sql query.
– code ninja
Sep 4 '13 at 7:17

There is not enough memory to run your script. PHP has reached the memory limit and stops executing it. This error is fatal, the script stops. The value of the memory limit can be configured either in the php.ini file or by using ini_set('memory_limit', '128 M'); in the script (which will overwrite the value defined in php.ini). The purpose of the memory limit is to prevent a single PHP script from gobbling up all the available memory and bringing the whole web server down.

php.ini

ini_set('memory_limit', '128 M');

php.ini

The first thing to do is to minimise the amount of memory your script needs. For instance, if you're reading a large file into a variable or are fetching many records from a database and are storing them all in an array, that may use a lot of memory. Change your code to instead read the file line by line or fetch database records one at a time without storing them all in memory. This does require a bit of a conceptual awareness of what's going on behind the scenes and when data is stored in memory vs. elsewhere.

If this error occurred when your script was not doing memory-intensive work, you need to check your code to see whether there is a memory leak. The memory_get_usage function is your friend.

memory_get_usage

Related Questions:

This error is most often encountered when attempting to reference an array value with a quoted key for interpolation inside a double-quoted string, when the entire complex variable construct is not enclosed in {}.

{}

This will result in Unexpected T_ENCAPSED_AND_WHITESPACE:

Unexpected T_ENCAPSED_AND_WHITESPACE

echo "This is a double-quoted string with a quoted array key in $array['key']"; //---------------------------------------------------------------------^^^^^

In a double-quoted string, PHP will permit array key strings to be used unquoted, and will not issue an E_NOTICE. So the above could be written as:

E_NOTICE

echo "This is a double-quoted string with an un-quoted array key in $array[key]"; //------------------------------------------------------------------------^^^^^

The entire complex array variable and key(s) can be enclosed in {}, in which case they should be quoted to avoid an E_NOTICE. The PHP documentation recommends this syntax for complex variables.

{}

E_NOTICE

echo "This is a double-quoted string with a quoted array key in {$array['key']}"; //--------------------------------------------------------------^^^^^^^^^^^^^^^ // Or a complex array property of an object: echo "This is a a double-quoted string with a complex {$object->property->array['key']}";

Of course, the alternative to any of the above is to concatenate the array variable in instead of interpolate it:

echo "This is a double-quoted string with an array variable " . $array['key'] . " concatenated inside."; //----------------------------------------------------------^^^^^^^^^^^^^^^^^^^^^

For reference, see the section on Variable Parsing in the PHP Strings manual page

The scope resolution operator is also called "Paamayim Nekudotayim" from the Hebrew פעמיים נקודתיים‎. which means "double colon" or "double dot twice".

This error typically happens if you inadvertently put :: in your code.

::

Related Questions:

Documentation:

The easiest way to trigger this error is running a()::b; or $a=::;.
– Ismael Miguel
Apr 13 '15 at 13:39

a()::b;

$a=::;

This means you're either using the same function/class name twice and need to rename one of them, or it is because you have used require or include where you should be using require_once or include_once.

require

include

require_once

include_once

When a class or a function is declared in PHP, it is immutable, and cannot later be declared with a new value.

Consider the following code:

class.php

<?php class MyClass { public function doSomething() { // do stuff here } }

index.php

<?php function do_stuff() { require 'class.php'; $obj = new MyClass; $obj->doSomething(); } do_stuff(); do_stuff();

The second call to do_stuff() will produce the error above. By changing require to require_once, we can be certain that the file that contains the definition of MyClass will only be loaded once, and the error will be avoided.

do_stuff()

require

require_once

MyClass

Is it worth mentioning using autoloading, and standards such as PSR-4 or even the now deprecated PSR-0 pretty much get rid of this by saving you from needing to use require/include yourself (bar a few bizarre edge cases).
– DanielM
Jun 1 '15 at 16:35

Happens when you try to use a variable that wasn't previously defined.

A typical example would be

foreach ($items as $item) { // do something with item $counter++; }

If you didn't define $counter before, the code above will trigger the notice.

$counter

The correct way would be to set the variable before using it, even if it's just an empty string like

$counter = 0; foreach ($items as $item) { // do something with item $counter++; }

Related Questions:

It happens when you call a file usually by include, require or fopen and PHP couldn't find the file or have not enough permission to load the file.

include

require

fopen

This can happen for a variety of reasons :

One common mistake is to not use an absolute path. This can be easily solved by using a full path or magic constants like __DIR__ or dirname(__FILE__):

__DIR__

dirname(__FILE__)

include __DIR__ . '/inc/globals.inc.php';

or:

require dirname(__FILE__) . '/inc/globals.inc.php';

Ensuring the right path is used is one step in troubleshooting these issues, this can also be related to non-existing files, rights of the filesystem preventing access or open basedir restrictions by PHP itself.

The best way to solve this problem quickly is to follow the troubleshooting checklist below.

Related Questions:

Related Errors:

or, in PHP 7.2 or later:

This notice occurs when a token is used in the code and appears to be a constant, but a constant by that name is not defined.

One of the most common causes of this notice is failure to quote a string used as an associative array key.

For example:

// Wrong echo $array[key]; // Right echo $array['key'];

Another common causes is a missing $ (dollar) sign in front of a variable name:

$

// Wrong echo varName; // Right echo $varName;

Or perhaps you have misspelled some other constant or keyword:

// Wrong $foo = fasle; // Right $foo = false;

It can also be a sign that a needed PHP extension or library is missing when you try to access a constant defined by that library.

Related Questions:

I would say the most common cause is forgetting $ in front of a variable, not arrays.
– Overv
Oct 10 '12 at 8:22

This warning can appear with various functions that are related to file and directory access. It warns about a configuration issue.

When it appears, it means that access has been forbidden to some files.

The warning itself does not break anything, but most often a script does not properly work if file-access is prevented.

The fix is normally to change the PHP configuration, the related setting is called open_basedir.

open_basedir

Sometimes the wrong file or directory names are used, the fix is then to use the right ones.

Related Questions:

This occurs most often on a shared host, people don't usually lock themselves out of directories :-)
– uınbɐɥs
Oct 7 '12 at 20:37

*

As the name indicates, such type of error occur, when you are most likely trying to iterate over, or find a value from array with a non-existing key.

Consider you, are trying to show every letter from $string

$string

$string = 'ABCD'; for ($i=0, $len = strlen($string); $i <= $len; $i++){ echo "$string[$i] n"; }

The above example will generate (online demo):

A B C D Notice: Uninitialized string offset: 4 in XXX on line X

And, as soon as the script finishes echoing D you'll get the error, because inside the for() loop, you have told PHP to show you the from first to fifth string character from 'ABCD' Which, exists, but since the loop starts to count from 0 and echoes D by the time it reaches to 4, it will throw an offset error.

D

for()

'ABCD'

0

D

4

Similar Errors:

Possible scenario

I can't seem to find where my code has went wrong. Here is my full error:

Parse error: syntax error, unexpected T_VARIABLE on line x

What I am trying

$sql = 'SELECT * FROM dealer WHERE id="'$id.'"';

Answer

Parse error: A problem with the syntax of your program, such as leaving a semicolon off of the end of a statement or, like the case above, missing the . operator. The interpreter stops running your program when it encounters a parse error.

.

In simple words this is a syntax error, meaning that there is something in your code stopping it from being parsed correctly and therefore running.

What you should do is check carefully at the lines around where the error is for any simple mistakes.

That error message means that in line x of the file, the PHP interpreter was expecting to see an open parenthesis but instead, it encountered something called T_VARIABLE. That T_VARIABLE thing is called a token. It's the PHP interpreter's way of expressing different fundamental parts of programs. When the interpreter reads in a program, it translates what you've written into a list of tokens. Wherever you put a variable in your program, there is aT_VARIABLE token in the interpreter's list.

T_VARIABLE

token

T_VARIABLE

Good read: List of Parser Tokens

So make sure you enable at least E_PARSE in your php.ini. Parse errors should not exist in production scripts.

E_PARSE

php.ini

I always recommended to add the following statement, while coding:

error_reporting(E_ALL);

PHP error reporting

Also a good idea to use an IDE which will let you know parse errors while typing. You can use:

Related Questions:

This error comes in two variatians:

$arr = [1, 2, 3];

This array initializer syntax was only introduced in PHP 5.4; it will raise a parser error on versions before that. If possible, upgrade your installation or use the old syntax:

$arr = array(1, 2, 3);

Related Questions:

Setting to 1 if it was 0 will stop the error but is this really any better than the suppression you said shouldn't be used (which I agree with)? I'd suggest most times it'd be likely some other message or value would be returned.
– James
Jun 17 at 0:56

For this example, if $var1 does == 0 then you can just set $var3 to $var2. Even if not doing that, the else is not needed at all as the assignment is the same in both cases so no else and assign outside the if
– James
Jun 17 at 0:58

$var1

$var3

$var2

if

Occurs when you try to call a non-static method on a class as it was static, and you also have the E_STRICT flag in your error_reporting() settings.

E_STRICT

error_reporting()

Example :

class HTML { public function br() { echo '<br>'; } }

HTML::br() or $html::br()

HTML::br()

$html::br()

You can actually avoid this error by not adding E_STRICT to error_reporting(), eg

E_STRICT

error_reporting()

error_reporting(E_ALL & ~E_STRICT);

since as for PHP 5.4.0 and above, E_STRICT is included in E_ALL [ref]. But that is not adviceable. The solution is to define your intended static function as actual static :

E_STRICT

E_ALL

static

public static function br() { echo '<br>'; }

or call the function conventionally :

$html = new HTML(); $html->br();

Related Questions:

Note that this also happens when TraitA::$x and TraitB::$x are the same value(say 'a') but TraitA::$a is public and TraitB::$a is private or protected
– Jelmergu
Nov 7 '17 at 18:45

Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

5C3HyrDn5J1afaII9rM6 fWSgcec n9uYeeBgax,bETn3mR

搜尋此網誌

Fjhtyj