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
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1楼 · 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

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