如何获取easygrid中的选定行值 [英] How to get selected row values in easygrid

问题描述

我在grails easygrid插件中创建自定义按钮open。我想在这个按钮中获得一些参数。如何从选定的行获取值？

 < grid：grid id ='jqgridinitial'name ='customerList'jqgrid .caption ='Customer'
 open =$ {g.createLink（controller：'customer'，action：'index'，params：[???]}
  
解决方案
这是返回所选行的javascript代码：
  var row = jQuery（＃jqgridinitial_table）。jqGrid（'getGridParam'，'selrow'）; 
  
但是，我认为您实际需要的是自定义jqgrid格式器：
  http://www.trirand.com/jqgridwiki/doku.php?id=wiki:custom_formatter  
 
 
 
你可以在这里看到一个例子： https://github.com/tudor-malene/Easygrid_example/blob/master/grail s-app / views / author / _jqgrid.gsp  
 
 
 
以下是它的工作原理（您点击作者名称并链接到wikipedia）： 
 199.231.186.169:8080/easygrid/author/list?impl=jqgrid 
 
 
 
在您的情况下，您应该根据实际行构建到客户的链接数据 
I create custom button "open" in grails easygrid plugin. I want get some parameters in this button. How I can get values from the selected row?
<grid:grid id='jqgridinitial' name='customerList' jqgrid.caption="'Customer'"
       open="${g.createLink(controller: 'customer', action: 'index', params:[???] }"

解决方案
This is the javascript code to return the selected row:
var row = jQuery("#jqgridinitial_table").jqGrid('getGridParam','selrow');
But, I think what you actually need is a custom jqgrid formatter:
http://www.trirand.com/jqgridwiki/doku.php?id=wiki:custom_formatter

You can see an example here: https://github.com/tudor-malene/Easygrid_example/blob/master/grails-app/views/author/_jqgrid.gsp

and here is how it works ( you click on the author name and it links to wikipedia):
    199.231.186.169:8080/easygrid/author/list?impl=jqgrid    

In your case you should construct the link to the customer based on the actual row data

                        
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