CODEWITHSUNDEEP
javascriptphpajaxjquery
arok
arok

Reputation: 1845

how to assign the retrieved data to the different td using ajax

When the user press button the data will be save into database.After success I want to retrieve data from database and place it in the proper td ( in the same row of the clicked td) i am success to retrieve data but not assign the retrieved data to the different td

ajax 

$(document).ready(function(){

        $('.edit2').on('click', function(){


         arr = $(this).attr('class').split( " " );
        var clientid=document.getElementById("client").value;

        $.ajax({    type: "POST",
        url:"clientnetworkpricelist/updateprice.php", 
        data: "value="+$('.ajax input').val()+"&rowid="+arr[2]+"&field="+arr[1]+"&clientid="+clientid,
        success: function(data){
                            $('#CPH_GridView1_clientprice'+arr[2]).empty();
                            $('#CPH_GridView1_clientprice'+arr[2]).append(data);                                                             
                            $('.ajax').html($(this).val());
                            $('.ajax').removeClass('ajax');
                                }});

                                  }  
                             );
                        });

HTML

<td  id="CPH_GridView1_clientprice'.$rows['net_id'].'" class="edit clientprice '.$rows["net_id"].'">'.$rows["clientprice"].'</td>
 <td  id="CPH_GridView1_Status'.$rows['net_id'].'" class="edit2 status '.$rows["net_id"].' "><img  src="image/'.$rows["status"].'f.png" /></td>

in my updateprice.php i connect to database and retrieve value from the database any just print the retrieve value like this 

print $newclientprice;

   print $status;

my result

both the value are showing now in the same td but i want it in separate td 0/01 in clientprice and increase in status 

Client Price      status
 0.01increase    |
                 |
                 |

any one help me thanks.

Upvotes: 2

Views: 204

Answers (1)

hanzo2001
hanzo2001

Reputation: 1398

updateprice.php

let's change this:

<?php
print $newclientprice;
print $status;

I'm not sure what you're trying to do. Because you are using jQuery, let's make use of JSON for our benefit. Important: the following code depends on no output being sent before it executes, and it will finish the php execution immediately

<?php
// set type so client can understand what was sent
header('Content-Type: application/json');
// transform our 2 values into a JSON array,
// this will be transparently transformed into an array for your ajax handler
echo json_encode(array($newclientprice, $status));
// end the script, this is what the client ajax request wanted
exit;

javascript.js

I took the liberty to rewrite your code so that it would seem clearer, and at least with a few comments too

$(document).ready(function(){
  var onClick, ajaxSuccessHandleMaker;
  onClick = function() {
    var
      url = 'clientnetworkpricelist/updateprice.php',
      clientid = $('#client')[0].value,
      classesArray = $(this).attr('class').split(" "),
      // send data as object, jQuery will transparently transform for the server
      data = {
        value : $('.ajax input').val,
        rowid : classesArray[2],
        field : classesArray[1],
        clientid : clientid
      };
    // send POST request and expect JSON
    $.post(url,data,ajaxSuccessHandleMaker(classesArray),'json');
  };
  // success returns the ajax handler with a closure on classesArray
  ajaxSuccessHandleMaker = function (arr) {
    // the handler EXPECTS an array, which is why we have to protect output in updateprice.php
    return function (arrayOf2vals) {
      $('#CPH_GridView1_clientprice'+arr[2]).html(arrayOf2vals[0]);
      $('#CPH_GridView1_Status'+arr[2]).html(arrayOf2vals[1]);
      // I am not sure what you want with the following 2 lines of code
      $('.ajax').html($(this).val());// what is this?
      $('.ajax').removeClass('ajax');// what is this?
    };
  };
  // set the onClick handler
  $('.edit2').click(onClick);
});

Finally

  • any questions are welcome
  • please test if it works, if it doesn't then comment back and I'll try to help out

Upvotes: 1

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