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How to display subgrid&

15/12/2008
05:55

SergeyK

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Forum Posts: 13

Member Since:
31/10/2008

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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'

16/12/2008
03:04

tony

Sofia, Bulgaria

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30/10/2007

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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

For professional UI suites for Java Script and PHP visit us at our commercial products site - guriddo.net - by the very same guys that created jqGrid.

16/12/2008
03:45

SergeyK

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Forum Posts: 13

Member Since:
31/10/2008

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

16/12/2008
04:23

tony

Sofia, Bulgaria

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30/10/2007

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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

For professional UI suites for Java Script and PHP visit us at our commercial products site - guriddo.net - by the very same guys that created jqGrid.

17/12/2008
03:26

SergeyK

Member

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Forum Posts: 13

Member Since:
31/10/2008

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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?

17/12/2008
11:28

tony

Sofia, Bulgaria

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Forum Posts: 7721

Member Since:
30/10/2007

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Hello,

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

Regards

Tony

For professional UI suites for Java Script and PHP visit us at our commercial products site - guriddo.net - by the very same guys that created jqGrid.

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