Ajax not sending var to url php - php

Question

i am trying to send the variables to server but url does not go to the php file.. What am i doing wrong? Is it because i am not using a submit button? Here is my code..
<table id="table">
<?php for($i=0; $i<$totalbesin; $i++) { ?>
<tr>
<td class="td" id="<?php echo "name".$i?>"><?php echo $besin[$i]['n']?></td>
<td class="td" id="<?php echo "kal".$i?>"><?php echo $besin[$i]['kal']?></td>
<td class="td" id="<?php echo "pro".$i?>"><?php echo $besin[$i]['pro']?></td>
<td class="td" id="<?php echo "kar".$i?>"><?php echo $besin[$i]['kar']?></td>
<td class="td" id="<?php echo "yag".$i?>"><?php echo $besin[$i]['yag']?></td>
<td class="ekle" id="yuk<?php echo $i?>">Ekle</td>
</tr>
<script type="text/javascript" >
$(function() {
$("#yuk<?php echo $i?>").click(function()
{
var name = $("#<?php echo "name".$i?>").val();
var kal = $("#<?php echo "kal".$i?>").val();
var pro = $("#<?php echo "pro".$i?>").val();
var kar = $("#<?php echo "kar".$i?>").val();
var yag = $("#<?php echo "yag".$i?>").val();
var dataString = '&name='+ name + '&kal=' + kal + '&pro=' + pro + '&kar=' + kar + '&yag=' + yag;
$.ajax({
type: "POST",
url: "dietup.php",
data: dataString,
cache: false,
success: function(html){
$("span#sonuc").text("<?php echo $i?>");
}
});
return false;
}); });
</script>
<?php } ?>
</table>

Answer 1

you need to change the val() to text(); like below as"td" tags not having an value tag
var name = $("#<?php echo "name".$i?>").text();
var kal = $("#<?php echo "kal".$i?>").text();
var pro = $("#<?php echo "pro".$i?>").text();
var kar = $("#<?php echo "kar".$i?>").text();
var yag = $("#<?php echo "yag".$i?>").text();

Answer 2

Take a look at ajaxForm & ajaxSubmit, otherwise you need a submit button.
http://malsup.com/jquery/form/

Answer 3

Try this
$.ajax({
type: "POST",
url: "dietup.php",
data: {
name: name,
kal: kal,
pro: pro
//and so on
},
cache: false,
success: function(html){
$("span#sonuc").text("<?php echo $i?>");
}
});

Answer 4

Use this to set the value into your variables:
var name = $("#name<?php echo$i?>").val();
var kal = $("#kal<?php echo $i?>").val();
var pro = $("#pro<?php echo $i?>").val();
var kar = $("#kar<?php echo $i?>").val();
var yag = $("#yag<?php echo $i?>").val();

Answer 5

In your code, as I understand, it seems that you are trying to assign values to your javscript variables using php.
There is a problem with syntax where you are assigning the values to your javscript variables.
Notice the following snippet from your code below:
var name = $("#<?php echo "name".$i?>").val();
var kal = $("#<?php echo "kal".$i?>").val();
var pro = $("#<?php echo "pro".$i?>").val();
var kar = $("#<?php echo "kar".$i?>").val();
var yag = $("#<?php echo "yag".$i?>").val();
Does anything seems not right?
Gotcha!!!! You are using double quotes to assign value in to your jquery selector and as well as for the php echo statement which is terminating your double quote.
Try using single quotes for the jquery selector and double quotes for the php echo statement as follows:
var name = $('#<?php echo "name".$i?>').val();

Answer 6

First of all, ajax parameter data must be object, not url string. If you want to send string url, add url string to url like this:
$.ajax({
url: "dietup.php?&name='+ name + '&kal=' + kal + '&pro=' + pro + '&kar=' + kar + '&yag=' + yag;"
})
But it will be wrong approach
I have divided yours script into two logical components.
First: is php loop generates the html table
Second: javascript function, who's sending ajax request
<table id="table">
<?php for($i=0; $i<$totalbesin; $i++) { ?>
<tr rel="<?php echo $i ?>">
<td class="td name" id="<?php echo "name".$i?>"><?php echo $besin[$i]['n']?></td>
<td class="td kal" id="<?php echo "kal".$i?>"><?php echo $besin[$i]['kal']?></td>
<td class="td pro" id="<?php echo "pro".$i?>"><?php echo $besin[$i]['pro']?></td>
<td class="td kar" id="<?php echo "kar".$i?>"><?php echo $besin[$i]['kar']?></td>
<td class="td yag" id="<?php echo "yag".$i?>"><?php echo $besin[$i]['yag']?></td>
<td class="ekle" id="yuk<?php echo $i?>">Ekle</td>
</tr>
<?php } ?>
</table>
<script type="text/javascript" >
$(function() {
$(".ekle").click(function()
{
var parent = $(this).closest('tr');
var data = {
name: parent.find('.name').val(),
kal: parent.find('.kal').val(),
pro: parent.find('.pro').val(),
kar: parent.find('.kar').val(),
yag: parent.find('.yag').val()
};
$.ajax({
type: "POST",
url: "dietup.php",
data: data,
cache: false,
success: function(html){
$("span#sonuc").text(parent.attr('rel'));
}
});
return false;
});
});
</script>

Answer 7

I found the problem.. It is because it does not get data from td, i tried getelementById but it did not work either.. Then i used
var name = '<?php echo $besin[$i]['n']?>';
and put the data in var like this.. Now it works.. Thank you for your time.. It would be better if i could get the data from td but its ok for now.. :)

Answer 8

I have a similar script, hope this help you:
<script type="text/javascript">
$(function() {
$("#yuk<?php echo $i?>").click(function(event)
{
var name = $("#<?php echo "name".$i?>").text();
var kal = $("#<?php echo "kal".$i?>").text();
var pro = $("#<?php echo "pro".$i?>").text();
var kar = $("#<?php echo "kar".$i?>").text();
var yag = $("#<?php echo "yag".$i?>").text();
var dataString = 'name='+ name + '&kal=' + kal + '&pro=' + pro + '&kar=' + kar + '&yag=' + yag;
//Uncomment to check the querystring
//alert("dietup.php?"+dataString);
$.ajax({
url:"dietup.php?"+dataString,
type:'GET',
dataType:'text',
cache:false,
success:function(data){
alert (data);
},
error:function(jxhr){
alert('Ops, error: '.jxhr.responseText);
}
}); //End Ajax
}); //End Click Function
}); //End Root Function
And here is the dietup.php example:
<?
//Get the parameters
$name = $_GET['name'];
//Do and return what you need
echo "your name is: ". $name;
?>