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Grid as Subgrid problem

30/11/2008
13:48

snip3r05

Portugal

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16/11/2008

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Hi.

I have this grid and subgrid

jQuery("#listsg11").jqGrid({
       url:'server.php?q=4',
    datatype: "json",
    height: 190,
       colNames:['Id', 'seccao', 'total','Data'],
       colModel:[
           {name:'id',index:'id', width:55},
           {name:'seccao',index:'seccao', width:100},
       ],
       rowNum:6,
       rowList:[8,10,20,30],
       imgpath: 'themes/sand/images',
       pager: jQuery('#pagersg11'),
       sortname: 'id',
    viewrecords: true,
    sortorder: "desc",
    multiselect: false,
    subGrid: true,
    caption: "Grid as Subgrid",
    subGridRowExpanded: function(subgrid_id, row_id) {
        var subgrid_table_id, pager_id;
        subgrid_table_id = subgrid_id+"_t";
        pager_id = "p_"+subgrid_table_id;
        rowdata = jQuery("#pagersg11").getRowData(row_id);
        seccao=rowdata['seccao'];
        alert(seccao);
    },
My problem is in the variable seccao that never as a value the alert always

shows "undefined".

Can u tell me what i am doing wrong?

Best Regards

Snip3r

30/11/2008
19:16

palobo

Portugal

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13/11/2008

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Hi, Please try the following and see if it wokrs:

seccao = rowdata.seccao;

30/11/2008
21:55

snip3r05

Portugal

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i still have “undefined” in the alert

01/12/2008
03:24

tony

Sofia, Bulgaria

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Hello,

Here is the error:

You have

rowdata = jQuery(”#pagersg11″).getRowData(row_id);

should be

rowdata = jQuery(”#listsg11″).getRowData(row_id);

The method should be applied to grid, not to pager

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.

01/12/2008
08:38

snip3r05

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Thanks a lot tony

Best Regards

Snip3r

01/12/2008
10:32

palobo

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NIce catch... missed that one 😉

Cheers;
P.

03/12/2008
16:41

snip3r05

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Hi Again!

I have a new probl :S, i have a third grid inside of the grid i has having the probl retreiving the data. now i need to get the data inside the second

grid into the the third but i dont now how to get the second grid name

so i can allter this->>>>"rowdata = jQuery(”secondgrid name ″).getRowData(row_id);""

i tryed passing the subgrid_table_id as argument like this:

subGridRowExpanded: function(subgrid_id, row_id,subgrid_table_id)

But the subgrid_table_id inside the expanded row is undified

Thanks in advanced

03/12/2008
16:50

snip3r05

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If it helps where is the code

   subGridRowExpanded: function(subgrid_id, row_id) {
        var subgrid_table_id, pager_id;
        subgrid_table_id = subgrid_id+"_t";
        pager_id = "p_"+subgrid_table_id;
        var rowdata = jQuery("#listsg11").getRowData(row_id);
        var data = rowdata.data;
        unidade = rowdata.unidade;
        alert(subgrid_table_id);
        $("#"+subgrid_id).html("<table id='"+subgrid_table_id+"' class='scroll'></table><div id='"+pager_id+"' class='scroll'></div>");
        jQuery("#"+subgrid_table_id).jqGrid({
            url:'server.php',
            datatype: "json",
            colNames:['Id','id_seccao','id_unidade', 'Periodo', 'Seccao', 'Total Horas', 'Horas Extra'],
           colModel:[
            {name:'id',index:'id', width:55,hidden:true},
            {name:'seccao',index:'seccao', width:55,hidden:true},
            {name:'unidade',index:'unidade', width:55,hidden:true},
               {name:'data',index:'data', width:100},
            {name:'seccao_desc',index:'seccao_desc', width:100},
               {name:'total',index:'total', width:80},
               {name:'extra',index:'extra', width:100}
            ],
            mtype: "POST",
            postData:{q:7,unidade:uni,permissoes:perm,data:data,seccao:seccao,user:user},
               rowNum:20,
               pager: pager_id,
               imgpath: 'themes/basic/images',
               sortname: 'seccao',
              sortorder: "asc",
              height: '100%',
            width: 650,
              subGrid: true,
            subGridRowExpanded: function(subgrid_id, row_id,subgrid_table_id) {
                var row = jQuery("#"+subgrid_table_id).getRowData(row_id);
                var data = row.data;
                unidade = row.seccao;
                alert(subgrid_table_id);
            },

04/12/2008
12:53

tony

Sofia, Bulgaria

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Hello,

maybe you should define variables before setting the grids

var subgrid, subsubgrid;

and set them accordantly, so that hey can be visible for

the grids

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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