Jquery easyui怎么得到datagrid里面的值和传到后台?

2022-11-20 01:25发布

var insertRows = $('#test').datagrid('getChanges','inserted');\x0d\x0a va

var insertRows = $('#test').datagrid('getChanges','inserted');\x0d\x0a va
1条回答
2022-11-20 01:59 .采纳回答

var insertRows = $('#test').datagrid('getChanges','inserted');\x0d\x0a var updateRows = $('#test').datagrid('getChanges','updated');\x0d\x0a var deleteRows = $('#test').datagrid('getChanges','deleted');\x0d\x0a var changesRows = {\x0d\x0a inserted : [],\x0d\x0a updated : [],\x0d\x0a deleted : [],\x0d\x0a };\x0d\x0a if (insertRows.length>0) {\x0d\x0a for (var i=0;i0) {\x0d\x0a for (var k=0;k0) {\x0d\x0a for (var j=0;j