AJAX doesn't get data from PHP file with Jquery

Question

I'm trying post data to PHP file but i can't receive any data from PHP file. Let me add codes. This is my jQuery function:

$(document).ready(function () {

$(function () {
$('a[class="some-class"]').click(function(){

   var somedata = $(this).attr("id");

   $.ajax({
      url: "foo.php", 
      type: "POST",
      data: "id=" + somedata,

      success: function(){
          $("#someid").html();
      },
      error:function(){
          alert("AJAX request was a failure");
      }
    });
    });
    });
});

This is my PHP file:

<?php
$data = $_POST['id'];

$con = mysqli_connect('localhost','root','','somedatabase');
if (!$con) {
  die('Could not connect: ' . mysqli_error($con));
}

mysqli_select_db($con,"database");
$sql="SELECT * FROM sometable WHERE id = '".$data."'";
$result = mysqli_query($con,$sql);


while($row = mysqli_fetch_array($result)) {

 echo $row['info'];

 }

mysqli_close($con);

?>

This what i have in HTML file:

<p id="someid"></p>

<a href="#page2" class="some-class" id="1">Data1</a>
<a href="#page2" class="some-class" id="2">Data2</a>

Note: This website is horizontal scrolling and shouldn't be refreshed. When i'm clicking links (like Data1) it's going to another page without getting data from PHP file

This is a POST request -.- do not use data: "id=" + somedata, but something like id:somedata — Marco Mura
– Marco Mura, Commented Dec 16, 2014 at 14:02
you need to add ? at ur url after .php url: "foo.php?", — Alex
– Alex, Commented Dec 16, 2014 at 14:03

jeroen · Accepted Answer · 2014-12-16 14:12:11Z

2

You have a few problems:

You are not using the data as mentioned in the other answers:
success: function(data){ $("#someid").html(data); },
You are not cancelling the default click action so your link will be followed:
$('a[class="some-class"]').click(function(e){ e.preventDefault(); ...;
As the id's are integers, you can use data: "id=" + somedata, although sending an object is safer in case somedata contains characters that need to be escaped:
data: {"id": somedata},;
You have an sql injection problem. You should cast the variable to an integer or use a prepared statement:
$data = (int) $_POST['id'];;
As also mentioned in another answer, you have two $(document).ready() functions, one wrapping the other. You only need one.

answered Dec 16, 2014 at 14:12

jeroen

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2 Comments

Korhan Yüzbaş Over a year ago

thank you! The only problem was forget to add 'data' to function and html. I want to ask another question for this, This function gets every data from PHP file but i want to take about 5 variable for each <p> How can i do that?

jeroen Over a year ago

@KorhanYüzbaş That's really another question, but in short, you could return json containing whatever you need and parse / use that in the success() function.

A. Rama · Accepted Answer · 2014-12-16 14:04:04Z

1

success: function(){
          $("#someid").html();
      },

should be:

success: function(data){
          $("#someid").html(data);
      },

answered Dec 16, 2014 at 14:04

A. Rama

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1 Comment

Korhan Yüzbaş Over a year ago

thank you so much it's working now! I want to ask another question for this, This function gets every data from PHP file but i want to take about 5 variable for each <p> How can i do that?

MH2K9 · Accepted Answer · 2014-12-16 14:04:13Z

1

You should add parameter in success

success: function(data){ //Added data parameter
      console.log(data);
      $("#someid").html(data);
  },

The data get the values what you echo in PHP end.

answered Dec 16, 2014 at 14:04

MH2K9

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Comments

Todd · Accepted Answer · 2014-12-16 14:06:35Z

1

This:

  success: function(data){
      $("#someid").html(data);
  },

and you have two document ready, so get rid of:

$(document).ready(function () { ...

});

answered Dec 16, 2014 at 14:06

Todd

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Comments

n0idea · Accepted Answer · 2014-12-16 14:06:42Z

0

data: "id=" + somedata,

Change it to:

data: { id : somedata }

answered Dec 16, 2014 at 14:06

n0idea

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2 Comments

Rory McCrossan Over a year ago

A querystring is perfectly valid for data

n0idea Over a year ago

somedata could be a not valid part of a querystring, so this solution would fix this problem.

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