TypeError: $.data(...) is undefined

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Topic: TypeError: $.data(...) is undefined (Read 8886 times)

Alfred

Full Member

Posts: 134

-Licensed User-

TypeError: $.data(...) is undefined

« on: October 21, 2019, 07:57:08 PM »

Whenever I call a function to open a dialog which has a datagrid, I got the following error in the console.

Code:

TypeError: $.data(...) is undefined

However this error does not cause any misbehaviour in my application. My code goes like this:

Code:

function openDialog(){
    $('body').desktop('openApp', {
         href:'datagridview.php',  
         //other code
    });
}

datagridview.php

Code:

<body>
    <table id="dg" data-options="href:view.php" class="easyui-datagrid">
         <th field="categoryName" halign="center" width="20" sortable="true" align="left">Category</th>
          //other codes follow
    </table>
</body>

I tested the code in Firefox, Chrome, Internet Explorer 11. I do not see any misbehavior. But since It is an error, I want to get rid of it. Please help.

Thanks
Alfred


« Last Edit: October 21, 2019, 08:39:35 PM by Alfred »	Logged

stworthy

Administrator
Hero Member

Posts: 3581

Re: TypeError: $.data(...) is undefined

« Reply #1 on: October 23, 2019, 07:57:09 PM »

Your 'datagridview.php' should be:

Code:

<table id="dg" data-options="fit:true" class="easyui-datagrid">
	<thead>
		<tr>
		<th field="categoryName" halign="center" width="200" sortable="true" align="left">Category</th>
		<th field="categoryName2" halign="center" width="200" sortable="true" align="left">Category2</th>
		</tr>
	</thead>
</table>

No error occurs on the console.


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Alfred

Full Member

Posts: 134

-Licensed User-

Re: TypeError: $.data(...) is undefined

« Reply #2 on: October 27, 2019, 09:44:29 PM »

Thanks for the reply.

the datagridview.php contains a a datagrid definion. It has a different url for getting the datagrid json. So the complete code looks like below:

This code is in index.php:

Code:

function openDialog(){
    $('body').desktop('openApp', {
         href:'datagridview.php',  
         //other code
    });
}

We have another file datagridview.php, which has the following datagrid code:

Code:

<table id="dg" data-options="url:view.php" class="easyui-datagrid">
         <th field="categoryName" halign="center" width="20" sortable="true" align="left">Category</th>
          //other codes follow
</table>

So when we run

Code:

openDiaglog()

function, It opens a new dialog. Inside the dialog, we have a datagrid that will load json from view.php. So the error still exists. The complete console error is:

TypeError: $.data(...) is undefined easyui.min.js:1:245666
   getPager http://test.com/js/easyui.min.js:1
   datagrid http://test.com/js/easyui.min.js:1
   loaded http://test.com/js/easyui.min.js:1
   jQuery 2
   loaded http://test.com/js/easyui.min.js:1
   datagrid http://test.com/js/easyui.min.js:1
   _78f http://test.com/js/easyui.min.js:1
   success http://test.com/js/easyui.min.js:6
   jQuery 4

I hope you can fix the error.

Thanks and regards,
Alfred


« Last Edit: October 27, 2019, 09:48:15 PM by Alfred »	Logged

stworthy

Administrator
Hero Member

Posts: 3581

Re: TypeError: $.data(...) is undefined

« Reply #3 on: October 28, 2019, 02:33:58 AM »

Please look at this example https://www.jeasyui.com/demo/test/test8.html. It works fine.

test8_1.php

Code:

<table id="dg" data-options="fit:true,url:'test8_2.php'" class="easyui-datagrid">
	<thead>
		<tr>
		<th field="categoryName" halign="center" width="200" sortable="true" align="left">Category</th>
		<th field="categoryName2" halign="center" width="200" sortable="true" align="left">Category2</th>
		</tr>
	</thead>
</table>

test8_2.php

Code:

<?php

$data = array(
	array('categoryName'=>'categoryName11','categoryName2'=>'categoryName12'),
	array('categoryName'=>'categoryName21','categoryName2'=>'categoryName22'),
	array('categoryName'=>'categoryName31','categoryName2'=>'categoryName32'),
);

echo json_encode($data);


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Alfred

Full Member

Posts: 134

-Licensed User-

Re: TypeError: $.data(...) is undefined

« Reply #4 on: October 28, 2019, 07:52:35 PM »

Thanks for the reply.

Your example just works fine, yet the error still exists in my attempted code. It may be due to the way the pagination works. In your example the pagination is not included. I am using jquery 1.12.4. and easyui 1.7.x.


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