如何删除行中的jqGrid与按钮的jqGrid - 删除一行按钮
http://jsfiddle.net/chandelyt/w4oudhh4/3/
HTML
<table id="list2">
<tr>
<td />
</tr>
</table>
<div id="jQGridDemoPager"></div>
JQuery的
var $j = $.noConflict(true);
var mydata = [
{ "UserName": "8125579231", "RoleId": 1, "Name": "John", "RoleName": "Administrator" },
{ "UserName": "9676078986", "RoleId": 1, "Name": "Mia", "RoleName": "Administrator" },
{ "UserName": "9703804807", "RoleId": 1, "Name": "Shan", "RoleName": "Administrator" },
{ "UserName": "9177458358", "RoleId": 1, "Name": "Tim", "RoleName": "Administrator" },
{ "UserName": "7760699118", "RoleId": 2, "Name": "Harry", "RoleName": "Sales" }
];
$j('#list2').jqGrid({
caption: "Employee Details",
data: mydata,
datatype: "local",
colNames: ["UserName", "RoleId", "Name", "RoleName","Delete"],
colModel: [
{ name: "UserName", index: 'UserName', width: 150 },
{ name: 'RoleId', index: "RoleId", width: 150 },
{ name: "Name", index: "Name", width: 150 },
{ name: "RoleName", index: "RoleName", width: 150 },
{ name: 'Delete', formatter: buttonFormatter }
],
rowNum: 10,
pager: '#jQGridDemoPager',
sortname: "UserName",
viewrecords: true,
sortorder: "desc",
});
function buttonFormatter(cellvalue, options, rowObject) {
return '<button type="button" onClick=clickFunction1();>Delete</button>';
};
function clickFunction1() {
var grid = $j('#list2');
var sel_id = grid.jqGrid('getGridParam', 'selrow');
var myCellData = grid.jqGrid('getCell', sel_id, 'Name');
alert("Selected Name: " + myCellData);
};
CSS
.ui-jqgrid .ui-state-highlight { background: silver; }
问题如下: 当我点击删除按钮(不选择行)这表明先前选定的行值(在这种情况下,名称)
我想点击任何按钮(和它应该选择该行 - 或不),并获取其字段值..我发布到服务器删除,然后刷新网格,一旦项目被删除。
它工作,如果我选择一行,然后单击按钮...我的工作的本地副本,但不能在小提琴......可能是becasue的旧版本的fiddeler
大。看起来不错 –