Print Page - datagrid calls form: how to pass id paramater

Title: datagrid calls form: how to pass id paramater
Post by: andyj on July 30, 2013, 02:12:13 AM

I want to call a form to edit a single record from a datagrid containing many records.

The edit form's datasource (data_select.php) dynamically queries a MySQL table to create a json file. It needs to receive the id key parameter.
e.g. data_select.php?entity_id=3 (where 3 is the parameter value being passed)

Code:

<?php
include '../../common/conn.php';
$eid = intval($_REQUEST['entity_id']);

$rs = mysql_query("SELECT entity_id, FirstName, MiddleName, LastName,Suffix,Phone,Mobile,AltPhone,Fax,Email FROM entity WHERE entity_id = $eid");
	
	$items = array();
	while($row = mysql_fetch_object($rs)){
		array_push($items, $row);
	}
	$result["row"] = $items;
	
	echo str_replace( array( '[', ']'), '', json_encode($items));
?>

The calling datagrid uses this function to open the edit form:

Code:

function openForm(){
   var row = $('#dg').datagrid('getSelected');
    if (row) {
        url = "../supplieredit/supplieredit.php?entity_id=" + row.entity_id;
        window.open(url,'_parent');
    }	
}

This seems to work fine

The edit form itself has its datasource definition like this:

Code:

$('#fm1').form('load','data_select.php?entity_id='+request.getParameter("entity_id"));

But it is not getting the parameter (herein lies the problem I think).

Can anyone help please? Thanks in advance :-\

Title: Re: datagrid calls form: how to pass id paramater
Post by: stworthy on July 30, 2013, 06:45:32 PM

Some syntax errors occur in your statement:

Code:

$('#fm1').form('load','data_select.php?entity_id='+request.getParameter("entity_id"));

If the 'entity_id' is generated from your php file, please use the code below.

Code:

$('#fm1').form('load','data_select.php?entity_id='+'<?php echo $eid;?>');

Title: Re: datagrid calls form: how to pass id paramater
Post by: andyj on July 31, 2013, 01:33:16 AM

Unfortunately that didn't do the trick.

Code:

$('#fm1').form('load','data_select.php?entity_id='+'<?php echo $eid;?>');

Doesn't get the parameter value of "entity_id" and add it to the url.
View Source shows a blank value, e.g. "supplieredit.php?entity_id="

The order of events is

Calling form opens window with &entity_id=[keyfieldvalue] - This works fine
Edit form is opened with the url including the parameter and parameter value, e.g. 'supplieredit.php?entity_id=6'
BUT the edit form is blank as the parameter is not being captured and passed to the data_select.php file that creates the json datasource
If you execute the url 'data_select.php?entity_id=6' in the browser it correctly filters the record where entity_id=6, so no apparent problems there
If you execute the url 'data_select.php?entity_id=' or 'data_select.php' in the browser it returns blank, which is what is happening

My only thoughts on why it is not working are:
May be the edit form's "load" method is firing before the url parameter is available to it?
Should I use the form's onBeforeLoad method to get the variable?

Is there any tutorial on how to pass variables via url parameters?
I guess I could use php session variables but it would be nice to be able to manipulate url parameters.
Thanks for your input, greatly appreciated.

Title: Re: datagrid calls form: how to pass id paramater
Post by: andyj on July 31, 2013, 01:45:42 AM

Got it now.

It should be:

Code:

$('#fm1').form('load','data_select.php?entity_id='+'<?php echo $_GET['entity_id'];?>');

Thanks for putting me on the right track.

EasyUI Forum

General Category => EasyUI for jQuery => Topic started by: andyj on July 30, 2013, 02:12:13 AM