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Ajax请求+JSON传回到页面数据的例子

2015-03-18 16:44 351 查看
前台页面(简略)

<span style="font-size:18px;"><span style="white-space:pre">		</span><tr>
<td height="32px" width="150px" align="center" bgcolor="#eaf8ff">机构编号</td>
<td>
<input type="text" name="" id="" size="40" onblur="getJgmc(this.value)"/>
<input type="hidden" name="nlyz.jdjg_id" id="jdjg_id"/>
</td>
</tr></span>
javaScrip代码段

function getJgmc(xkzh){
$.ajax({
method : "POST",
url : "jdjg/jdjg_getJgmcByXkzh.do",
data : "xkzh="+xkzh,
dataType : "json",
cache : false,

error : function() {
alert("请求失败");
},

success : function(data) {
var jgmc = data.jgmc;
var jdjg_id = data.jdjg_id;
var xzqy = data.xzqy;
if(jgmc == "null"){
alert("无对应机构,请核准机构编号!");
}else{
document.getElementById("jgmc").value = jgmc;
document.getElementById("jdjg_id").value = jdjg_id;
document.getElementById("xzqy").value = xzqy;
}
}
});


后台数据传输实现

public void getJgmcByXkzh() throws IOException {
jdjg = jdjgService.getJgmcByXkzh(request.getParameter("xkzh"));
response.setCharacterEncoding("utf-8");
PrintWriter out = response.getWriter();
JSONObject jo = new JSONObject();
if (jdjg!=null) {
jo.put("jgmc", jdjg.getJg_jgmc());
jo.put("jdjg_id", jdjg.getId());
jo.put("xzqy", jdjg.getJg_xzqy());
}else {
jo.put("jgmc", "null");
}
out.print(jo);
out.flush();
out.close();
}
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