jQuery Grid Plugin - jqGridForum

11/11/2009
06:25

jmiguel.rodriguez

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Hello all,

Althoght I'm a long time java programmer, I've just started with javascript, jquery and jqgrid.

I'm trying to implement jqgrid in one application and I'm having some problems. Sure it will be an stupidity but I've been some days trying to find by myself what's happening and cannot find it.

This is my first grid (code below). Html page loads ok, and in firefox everything works fine, but not in IExplorer neither GoogleChrome. In IE/chrome I get the jqgrid correctly displayed but is not filled with data. I see the call to server (java servlet) in my logs and the xml output data is served back to jqgrid/jquery, but never displayed.

I've tried to server the xml with different content types (I guess it must be "Content-type: text/xml;charset=utf-8" , or, in my case ISO-8859-15) but I've also tried with text/html without success.

The error I see in IExplorer (v6 and v8) is an 'Object required' in jquery-1.3.2.min.js line 12 char 7209 , and in that point I see:

return E.nodeType===9&&E.documentElement.nodeName!=="HTML"||!!E.ownerDocument&&o.isXMLDoc(E.ownerDocument)

The code: html/javascript:

-----------------------------------------------------------------------------------------------------------------------------

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"

"http://www.w3.org/TR/html4/loose.dtd">

<html>

<head>

<title>jqGrid test</title>

jQuery().ready(function(){

jQuery("#list").jqGrid({

url:'/scanciax/servlet/lwasScancia?vsPage=jqdataPersonas',

datatype: "xml",

mtype: 'GET',

colNames:['id','Nombre','Apellido 1', 'Apellido 2', 'Tratamiento'],

colModel :[

{name:'idpersona', index:'idpersona', width:1,hidden:true},

{name:'snombre', index:'snombre', width:155, align:'left' , sortable:true},

{name:'sapellido1', index:'sapellido1', width:90, align:'right', sortable:true},

{name:'sapellido2', index:'sapellido2', width:380, align:'right', sortable:true},

{name:'idtratamiento', index:'idtratamiento', width:10, align:'right', sortable:false }

],

pager: jQuery('#pager'),

rowNum:10,

rowList:[10,20,50,100,250,500],

sortname: 'sapellido1',

sortorder: "asc",

imgpath: 'themes/basic/images',

caption: 'My test grid'

});

</script>

</head>

<body>

</body>

</html>

-----------------------------------------------------------------------------------------------------------------------------

Sample XML data served by the url "/scanciax/servlet/lwasScancia?vsPage=jqdataPersonas" is as follows:

-----------------------------------------------------------------------------------------------------------------------------

<?xml version="1.0" encoding="ISO-8859-15"?>

<rows>

<cell>- OTTO FASSBENDER</cell>

</row>

<cell>SERVICIO</cell>

</row>

.......

</rows>

-----------------------------------------------------------------------------------------------------------------------

Any help will be highly appreciated. Thank you in advanced .

12/11/2009
10:47

jmiguel.rodriguez

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I reply yo myself as I've just find what (stupid of me, I knew) was wrong:

I was setting Content-type on xml data from server as 'Content-type: text/xml', no just 'text/xml' as I should. The great thing is firefox doesn't complained about it.

Thank you all anyway

09/01/2010
02:37

Jeff N.

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I'm having the exact same problem as you and I don't quite understand how you fixed the xml error. My code for my php code that builds the XML data looks like:

. . .

        // we should set the appropriate header information
        if ( stristr($_SERVER["HTTP_ACCEPT"],"application/xhtml+xml") )
        {
            header("Content-type: application/xhtml+xml;charset=utf-8");
        } else {
            header("Content-type: text/xml;charset=utf-8");
        }

        echo "<?xml version='1.0' encoding='utf-8'?>";
        echo "<rows>";
        echo "<page>".$page."</page>";
        echo "<total>".$total_pages."</total>";
        echo "<records>".$rowCt."</records>";

        foreach ($result as $row)
        {
            echo "<row id='".$row->PART_ID."'>";
            echo "<cell><![CDATA[".$row->PART_ID."]]></cell>";
            echo "<cell><![CDATA[".$row->EAU."]]></cell>";
            echo "<cell><![CDATA[".$row->MINPER."]]></cell>";
            echo "<cell><![CDATA[".$row->NOTIFY."]]></cell>";
            echo "<cell><![CDATA[".$row->STATUS."]]></cell>";
            echo "</row>";
        }

        echo "</rows>";

. . .

Do you have any suggestions?

09/01/2010
10:38

olaf

GRENOBLE (FRANCE)

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I would write : pager: '#pager'

and not : pager: jQuery('#pager')

11/01/2010
18:36

Jeff N.

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olaf said:

I would write : pager: '#pager'

and not : pager: jQuery('#pager')

Are you talking about my code or the post above? Are there any ideas out there for the IE error? Things work great in firefox.

13/01/2010
18:19

tony

Sofia, Bulgaria

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Hello,

Jeff, What does not work exactley for you?

Best 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.