Question

ajax json alert 是object，怎么查看data里面的数据是不是 要的

Answer 1

1、首先前台用Ajax，其中注意dataType一定要选择json方式，Action成功返回给页面的Json内容是这样的[{"number":"V006","names":"LiLei"}],可见comment['names']对应"names":"LiLei"，comment['number']对应"number":"V006"。
$.ajax({
type: "post",
url:'apply/mystudent.action?',
cache: false,
dataType : "json",
success: function(data){

$.each(data, function(commentIndex, comment){

alert("姓名"+ comment['names']);

alert("学号"+comment['number']);

});
}
});

2、Ajax的URL指向在java的action中mystudent方法，返回的list其实是一个对象Student，包括了names和nunmber字段
public String mystudent() throws Exception{
List list=priceService.query();//调用接口实现类

this.jsonUtil(list);

return null;
}

3、action页面专门写一个方法jsonUtil来做为json方法
// 调用json工具方法，传入参数alist
public void jsonUtil(Object accountlist) throws Exception {
HttpServletResponse response = ServletActionContext.getResponse();
log.info("JSON格式：" + accountlist.toString());

String returnJson = JsonConvert.returnJson(accountlist);
response.setCharacterEncoding("utf-8");
response.getWriter().println(returnJson);
}

4、我用的是一种比较新的json包jackson
import java.io.StringWriter;

import org.codehaus.jackson.map.ObjectMapper;

public class JsonConvert {
static String jsonStr;
public static String returnJson(Object object) throws Exception{
ObjectMapper objectMapper = new ObjectMapper();
StringWriter stringWriter = new StringWriter();
objectMapper.writeValue(stringWriter, object);

jsonStr = stringWriter.toString();
return jsonStr;
}
}

Answer 2

在ajax success中用eval转换返回的对象然后alert

或者打开firebug看控制台ajax请求返回的HTML是什么

Answer 3

alert 是object 说明返回没成功。要检查下程序哪里有错误