jQuery Grid Plugin - jqGrid - Topic: How to display subgrid&

tony on How to display subgrid&

Wed, 17 Dec 2008 11:28:27 +0200

Hello,

look at the demo page json subgrid. Maybe you miss here to obtain the parameter using $_GET -

Regards

Tony

SergeyK on How to display subgrid&

Wed, 17 Dec 2008 03:26:17 +0200

Hi,tony

thanks,

i recived data from subgrid.php, but it didn't displayed yet.

my subgrid.php

$db = mysql_connect("localhost", "log", "pas")
or die("Connection Error: " . mysql_error());
mysql_select_db("base") or die("Error conecting to db.");

$SQL = "SELECT Code,Number,Name FROM tabSub WHERE Code = '".$Code."'";
$result = mysql_query( $SQL ) or die("Couldn't execute query.".mysql_error());

$responce->page = 1;
$responce->total = 1;
$responce->records = 1;
$i=0;
while($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
     $responce->rows[$i]['id']=$row[Number];
     $responce->rows[$i]['cell']=array($row[Number],$row[Name]);
     $i++;
}

echo $json_encode($responce);

my main.php

<body>
<script type='text/javascript'>
<!--
jQuery(document).ready(function(){
jQuery('#list').jqGrid({
url:'DataSalary2.php?q=1',
datatype: 'json',
mtype: 'GET',
colNames:['ÐÐ¾Ð¼ÐµÑ€','ÐšÐ¾Ð´'],
height: '100%',
colModel :[
{name:'Number',index:'Number', width:100, align:'center',search:true, jsonmap: 'Number' , sorttype:'int'},
{name:'Code',index:'Code', width:100,align:'left', search:true, jsonmap: 'Code' ,sorttype:'text'}],
imgpath: 'themes/green/images',

rowNum:10,
rowList:[10,20,30],
viewrecords: true,
subGrid : true,
subGridUrl: 'subgrid.php?nd='+new Date().getTime(),
subGridModel: [{ name : ['Number','Name'],
width : [55,200],
params:['Code']} ],
pager: jQuery('#pager'),
caption: 'S'
}).navGrid('#pager',{edit:false,add:false,search:false,del:false});
});
//-->
</script>
<table id='list' class='scroll' cellpadding='0' cellspacing='0'></table>
<div id='pager' class='scroll' style='text-align:center;'></div>

</body>");

what is wrong?

tony on How to display subgrid&

Tue, 16 Dec 2008 04:23:06 +0200

Hello,

In order to have something in the subgrid, you should configure

server.php to return the needed data. Read the docs how to do that.

Regards

Tony

SergeyK on How to display subgrid&

Tue, 16 Dec 2008 03:45:59 +0200

i haven't response from server. subgrid.php hasn't recived any data.

tony on How to display subgrid&

Tue, 16 Dec 2008 03:04:25 +0200

Hello,

Did you have respnse from the server when you click on the plus.

Also what data return the script in subgrid.php

Check your query for the subgrid.

Also look at the examples how to configure.

Regards

Tony

SergeyK on How to display subgrid&

Mon, 15 Dec 2008 05:55:39 +0200

when i pressed '+', there no any rows of subgrid

i think, that it hasn't entered to 'subgrid.php' at all.

What else should i wright to display subgrid?

jQuery('#list').jqGrid({
url:'DataSalary2.php?q=1',
datatype: 'json',
mtype: 'GET',
colNames:['ÐÐ¾Ð¼ÐµÑ€','ÐšÐ¾Ð´'],
height: '100%',
colModel :[
{name:'Number',index:'Number', width:100, align:'center',search:true, jsonmap: 'Number' , sorttype:'int'},
{name:'Code',index:'Code', width:100,align:'left', search:true, jsonmap: 'Code' ,sorttype:'text'}],
imgpath: 'themes/green/images',
jsonReader: {
                repeatitems: true,
                cell: 'cell',
           id: 'id'
            },
subGrid : true,
subGridUrl: 'subgrid.php',
subGridModel: [{ name : ['Number','Name'],
                    width : [55,200],
     params:['Code']} ],
pager: jQuery('#pager'),
rowNum:10,
rowList:[10,20,30],
viewrecords: true,
sortname: 'Team'