jqGrid tableToGrid“options”参数 [英] jqGrid tableToGrid "options" parameter

问题描述

基础

您好，我看到tableToGrid方法（由Peter Romianowski定义） tableToGrid选择器，选项）在jqGrid维基上，但是找不到任何有选项的文档

Hi all, I have see the tableToGrid method (by Peter Romianowski) defined as tableToGrid(selector, options) on the jqGrid wiki, but cannot find anywhere that has documentation of the options

有没有人知道这些或在哪里找到他们？编辑：感谢Oleg，解决了！

Does anyone know about these or where to find them? Thanks Oleg, resolved!

更多

do以形式包装生成的jqGrid，它将提交表中列中的复选框值。我的问题是 tableToGrid 方法似乎是从复选框中删除 name / code>元素，因此它们不会以表单提交。

What Im actually trying to do is encase the resulting jqGrid in a form, which will submit the checkbox values that are in a column of the table. My problem is that the tableToGrid method appears to be removing the name property from the checkbox elements and hence they are not being submitted with the form post.

推荐答案

RESOLVED

RESOLVED

jqGrid tableToGrid方法将在原始表中找到复选框的值，并自动实现 multiSelect：true 选项，显示内在的复选框。要获取所选行ID的列表，只需调用

The jqGrid tableToGrid method will find the value of the checkboxes in the original table and automatically implement the multiSelect: true option on the resulting jqGrid, showing intrinsic checkboxes. To get a list of the selected rows IDs, simply call

$('#grid').getGridParam('selarrrow')

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