EasyUI Forum

General Category => EasyUI for jQuery => Topic started by: sreekuttan on May 27, 2014, 09:28:43 PM

Title: Server Response from destroyUrl / updateUrl
Post by: sreekuttan on May 27, 2014, 09:28:43 PM

Consider the following sample code

Code:

$('#tt').edatagrid({
    url: 'datagrid_data.json',
    saveUrl: 'savedetails.php',
    updateUrl: 'upddetails.php',
    destroyUrl: 'delrow.php'
});

and server side

Code:

//savedetails.php
<?php
extract($_GET);
extract($_POST);
$query = "insert into tab values ('$a1','$a2','$a3')";
mysql_query($query);
echo "Row inserted";
?>

How I can get the response of

Code:

"Row inserted"

at the front end code on addition of a new row?

Title: Re: Server Response from destroyUrl / updateUrl
Post by: stworthy on May 28, 2014, 08:21:53 AM

Your 'savedetails.php' must return the saved row. When a new row saved successfully, the 'onSave' event fires, in which you will know what row has been saved.

Code:

$('#tt').edatagrid({
    url: 'datagrid_data.json',
    saveUrl: 'savedetails.php',
    updateUrl: 'upddetails.php',
    destroyUrl: 'delrow.php',
    onSave: function(index,row){...}
});