Showing the data of the right side of the TimeLine
How can I show the description Data on the left or the right side of the timeline acording of it Candidate.Saving result?
If it's Candidate.Saving == true it should be on the left, if it's Candidate.Saving==false it should be on the right side of the time line.
I will shou here my candidate Model And my View.
Candidate.Saving == true
Candidate.Saving==false
public class Candidate : BaseEntity
{
public int Id { get; set; }
public string Name { get; set; }
public int Number { get; set; }
public string ProfileText { get; set; }
public Byte CV { get; set; }
public string CVNAME { get; set; }
public List<Profile> ProfileList { get; set; }
public String Description { get; set; }
public Boolean Saving { get; set; }
public string Title { get; set; }
public DateTime DateOfDescription { get; set; }
}
Here is my view:
@model HCCBPOHR.Data.Candidate
@{
ViewData["Title"] = "CandidateHistory";
}
<h2>Canidate - @Model.Name</h2>
<label>History</label>
<hr />
History Of @Model.Name
@foreach (var Description in Model.Description)
{
@if (Model.Saving == true)
{
<ul>
<li class="timeline">
Title
@Model.DateOfDescription.ToShortDateString()
</div>
</li>
</ul>
}
@if (Model.Saving == false)
{
<ul>
<li class="timeline-inverted">
Title
@Model.DateOfDescription.ToShortDateString()
</div>
</li>
</ul>
}
}
</div>
@*
Isto é para inverter o lado
Title
@Model.DateOfDescription.ToShortDateString()
</div>
</li>
</ul>
</div>
</div>*@
<button type="button" class="btn btn-primary" data-toggle="modal" data-target="#exampleModal" data-whatever="@Model.Saving"> Add History Description</button>
Selects
Candidate
Hitachis
</form>
</div>
</div>
</div>
</div>
What I've tried to do:
I have tried to make a foreach method where the side depends of the Candidate.Saving, But the result was this:

I have managed to fix it.
This is how I did It:
History Of @Model.Name
@if (Model.Saving == true)
{
Title
@Model.DateOfDescription.ToShortDateString()
</div>
</li>
</ul>
}
@if (Model.Saving == false)
{
<ul class="timeline">
<li class="timeline-inverted">
Title
@Model.DateOfDescription.ToShortDateString()
</div>
</li>
</ul>
}
</div>
</div>
I only have on doubt right now, to be able to add more than one Description for the candidate I should Create a list off Descriptions
and do a foreach method in the view right?
Right now my controller looks like this:
public IActionResult CandidateHistory(int Id)
{
using (var applicationcontext = new ApplicationContext())
{
var candidate = applicationcontext.Candidates.Where(s => s.Id.Equals(Id)).SingleOrDefault();
if (candidate == null)
{
return NotFound();
}
Candidate candidates = new Candidate();
candidates.DescriptionList.Add(candidate);
candidates.DescriptionList.Add(new Candidate { Description = candidate.Description });
candidates.DescriptionList.Add(new Candidate { Title = candidate.Title });
candidates.DescriptionList.Add(new Candidate { DateOfDescription = candidate.DateOfDescription });
candidates.DescriptionList.Add(new Candidate { Saving = candidate.Saving });
applicationcontext.Candidates.Add(candidates);
return View();
}
}
[HttpPost, ActionName("CandidateHistory")]
[ValidateAntiForgeryToken]
public async Task<IActionResult> CandidateHistoryPost(int? id, Candidate candidatelist)
{
if (id == null)
{
return NotFound();
}
using (var context = new ApplicationContext())
{
var candidateUpdate = await context.Candidates.SingleOrDefaultAsync(s => s.Id == id);
if (await TryUpdateModelAsync<Candidate>(candidateUpdate, "", s => s.Description, s => s.Title, s => s.DateOfDescription, s => s.Saving))
{
try
{
await context.SaveChangesAsync();
return RedirectToAction(nameof(CandidateHistory));
}
catch (DbUpdateException /* ex */)
{
//Log the error (uncomment ex variable name and write a log.)
ModelState.AddModelError("", "Unable to save changes. " +
"Try again, and if the problem persists, " +
"see your system administrator.");
}
}
return View(candidateUpdate);
}
}
I'm trying to output an Candidate.DescriptionList to the CandidateHistory View so that way I can save into that list all the descriptions, description title, date and "saving", so i can do a foreach method to output all the description of that user.
But now I have this error:
System.NullReferenceException: 'Object reference not set to an instance of an object.
HCCBPOHR.Data.Candidate.DescriptionList.get returned null.`
1 Answer
1
I have managed to fix it.
This is how I did It:
History Of @Model.Name
@if (Model.Saving == true)
{
Title
@Model.DateOfDescription.ToShortDateString()
</div>
</li>
</ul>
}
@if (Model.Saving == false)
{
<ul class="timeline">
<li class="timeline-inverted">
Title
@Model.DateOfDescription.ToShortDateString()
</div>
</li>
</ul>
}
</div>
</div>
I only have on doubt right now, to be able to add more than one Description for the candidate I should Create a list off Descriptions
and do a foreach method in the view right?
By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.
Popular posts from this blog
JMeter fails on beanshell imports I had been using Beanshell pre-processor to add custom cookies created through javascript. While it used to work perfectly earlier, with the same line of codes I started to recieve errors in my pre-processor. ERROR - jmeter.util.BeanShellInterpreter: Error invoking bsh method: eval Sourced file: inline evaluation of: ``import org.apache.jmeter.protocol.http.control.CookieManager; import org.apache. . . . '' : TargetError WARN - jmeter.modifiers.BeanShellPreProcessor: Problem in BeanShell script org.apache.jorphan.util.JMeterException: Error invoking bsh method: eval Sourced file: inline evaluation of: ``import org.apache.jmeter.protocol.http.control.CookieManager; import org.apache. . . . '' : TargetError The lines of code is as per suggestion provided in https://javaworks.wordpress.com/2011/08/05/setting-cookie-in-jmeter/ import org.apache.jmeter.protocol.http.control.CookieManager; import org.apache.jmeter.protocol.http.control....
Why in node-red my HTTP POST no receive payload from inject? i try use https://rpinotify.it/invio-di-testo.html i want send message with http post but i can't, see images i use Node-Red to fred.sensetecnic, i try use json, strings... i can't undestrand, can you help me? 3/7/2018, 08:43:15node: 9ee1f8ea.e1c5b8 msg.payload : Object object text: "testo" 3/7/2018, 08:43:15node: 9ee1f8ea.e1c5b8 msg.payload : Object object info: object datetime: "2018-07-03 06:43:15" ip: "54.149.86.44, 108.162.246.239" message: "'text' or 'img' POST parameter are required" response: 400 [ { "id": "e6ce432b.5361", "type": "tab", "label": "Flow 2", "disabled": false, "info": "" }, { "id": "1b6aa9ad.127af6", "type": "inject", "z": ...
PHP contact form sending but not receiving emails I have created a contact form using PHP to send the data. The form seems to send fine as there are no errors and success message appears, however I am not receving the emails into the designated inbox. I am using PHPMailer as I originally tried just using the php 'mail' command which I now understand is a bit hit and miss. I cannot see why I would not be receving the emails so I would be very grateful for any help that could be given. I'm quite new to PHP so please be patient with me :) <?php use PHPMailerPHPMailerPHPMailer; use PHPMailerPHPMailerException; $msg = ""; if (isset($_POST['submit'])) { require 'phpmailer/src/Exception.php'; require 'phpmailer/src/PHPMailer.php'; require 'phpmailer/src/SMTP.php'; function sendemail ($to, $from, $body) { $mail = new PHPMailer(true); $mail->setFrom($from); $mail->addAddress($to); $ma...
If you’ve managed to fix it, you should post an answer detailing your fix, then accept that answer.
– Cole Johnson
Jul 2 at 13:45