Php Jqgrid将选择行列值传递给dataurl

我试图将列值传递给php，并根据列值在编辑表单上加载选择项

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

JS

可以使用定义为函数see和的dataUrl。函数dataUrl get 3参数第一个是rowid，第二个是来自当前列的值-在您的情况下为Local_Description，这将被初始化为网格的DOM，因此您可以使用$this.jqGridgetRowData，rowid或$this.jqGridgetCell，rowid，Description。通过这种方式，您可以生成所需的任何dataUrl值

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'} 
}