Jqgrid通过选择列行值到dataurl

2014-07-26 51 views 0 likes

我想通过列值到php加载基于列值的编辑表单上的选择项目。Jqgrid通过选择列行值到dataurl

dataUrl:'includes/Opera_classif.php?Op=local&id=' +ID +Description +id_local

mtype: 'GET', 
colNames: [ "ID","Descrição","ID Local", "Local","Select Local"], 
colModel: [ 
{name:'ID',index:'ID', width:20, sorttype:"int"}, 
{name:'Description',index:'Description', width:150, editable: true,editrules:{required:true}}, 
{name:'id_local',index:'id_local',hidden:true, width:20, editable: true,editrules:{required:true}}, 
{name:'Local_Description',index:'Local_Description', width:100, editable: true,editrules:{required:true}}, 
{ 
    name:'escolhe_local',index:'escolhe_local', width:80,resizable:true, hidden:true, editrules:{edithidden: true }, 
    align:"left",sorttype:"text",editable:true,edittype:"select", 
    editoptions:{dataUrl:'includes/Opera_classif.php?Op=local'} 
}

来源

2014-07-26 Jorge

回答

人们可以使用定义为函数dataUrl（见here和here）。函数dataUrl得到3个参数（第一个是rowid，第二个是来自当前列的值 - Local_Description）和this将被初始化为网格的DOM（因此您可以使用$(this).jqGrid("getRowData", rowid)或$(this).jqGrid("getCell", rowid, "Description")举例）。通过这种方式，您可以生成您需要的任何dataUrl值。

来源

2014-07-27 13:40:04 Oleg

Jqgrid通过选择列行值到dataurl

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