jQuery ajax - .post() problem

[kirill578]

This code doesn't work/execute anything ( ajax_request.php works fine)

 

    $.post("ajax_request.php", { vote: "a", id: "2", category: "0" },
      function(data) {
      $("div#result_a_show").text("" + data.a);
      $("div#result_b_show").text("" + data.b);
      $('a#next_link').attr('href', data.link);
      }, "json");   

[RichardRotterdam]

I think you're using the function incorrect.

Try something like:

$.post({
  url: 'yoururlhere.php',
  data: {data: 'here'}
  dataType: 'json',
  success: function(data) {
    // do stuff here when the data returns
  }
});

[kirill578]

 

i tried this. but now all the code doesn't working.

 

    $.post({
      url: 'ajax_request.php',
      data: { vote: 'b', id: '2', category: '0' }
      dataType: 'json',
      success: function(data) {
        $("div#result_a_show").text("" + data.a);
        $("div#result_b_show").text("" + data.b);
        $('a#next_link').attr('href', data.link);
      }
      });

[RichardRotterdam]

you forgot a comma after the data line

 data: { vote: 'b', id: '2', category: '0' },// comma here

 

Have you got something like firebug it's a lot easier to see what you have done wrong if you get some sort off error feedback

[kirill578]

Now i have the same problem, it doesn't execute anything

 

Can you look at the code here?

 

http://maadif.co.cc/ajax/prefer/index.php

[RichardRotterdam]

It does do something it doesn't find the url. It's because jQuery.post() and jQuery.ajax() work differently.

Use firebug to see what's going on.

If you change $.post to $.ajax and add type:'post as parameter it should prob work.

 

$.ajax({
    type:'post'
});

[kirill578]

I downloaded FireBug. Is says that the POST send successful and it gets the right response but the problem has I already said before is that it doesn't execute the success function. I tried to add alert into it but i doesn't apper as well. ( My feeling that jQuery hates me  :'( )

[RichardRotterdam]

What I suggested works fine for me. what does your code look like now?

Edit I took a look at the link and noticed the change try a console.log and see if something happens. or do a

alert('some text') 

that way you'll be sure the function runs

[kirill578]

That is exactly what I did. Just look at the code here

 

http://maadif.co.cc/ajax/prefer/index.php

[kirill578]

I have finally found out that the problem is the PHP file 

its output starts in the third line..

[cssfreakie]

I have finally found out that the problem is the PHP file 

its output starts in the third line..

 

could you maybe show what you mean by that, and what was the solution. sorry for asking but i am pretty eager to learn this

[haku]

It probably means that the output file was printing three linebreaks before executing it's code.

[cssfreakie]

It probably means that the output file was printing three linebreaks before executing it's code.

ok cool thanks! haku