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动态的二级联动步骤

发布时间: 2012-11-26 11:48:49 作者: rapoo

动态的二级联动方法

第二个下拉框的内容根据第一个选择的不同而不同。


前台代码:

public void AjaxGetLicenseInfo() throws Exception {HttpServletResponse response = ServletActionContext.getResponse();response.setCharacterEncoding("UTF-8");String[] str = new String[] { "0002%" };assessSortList = mouthassessService.searchSortinfo(str, SortListValue);String json = "{LicenseList:[";for (int i = 0; i < assessSortList.size(); i++) {String point = ",";if (i == assessSortList.size() - 1)point = "";json += "{LICENSEID:'"+ ((Points) assessSortList.get(i)).getPointsID()+ "',LICENSENAME:'"+ ((Points) assessSortList.get(i)).getPointsName() + "'}"+ point;}json += "]}";response.getWriter().print(json);}

JSON字符串要对应,

要导入 jquery-1.8.3.min.js

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