spring mvc ajax form提交报请求因 HTTP 状态 415 失败: Unsu...

最新推荐文章于 2023-09-22 11:09:07 发布

转载最新推荐文章于 2023-09-22 11:09:07 发布 · 281 阅读

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原文链接：https://my.oschina.net/stonezing/blog/122265

文章标签：

#java #json #测试

本文探讨了使用Ajax GET方式与jQuery Form提交数据时遇到415错误的问题，并给出了具体的解决方案，包括调整Controller处理方法及参数接收方式。

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之前用ajax异步GET方式提交数据

$.ajax({
      type: "GET",
      contentType: "application/json; charset=utf-8",
      dataType: "json",
      url: "<%=request.getContextPath()%>/test.html",
      data:{username:"zsd"},
      success: function(text){
       alert(text.username)
      },
      error: function(err) {
                           alert("err");
                       }
     });

Controller 中-------------------

@RequestMapping(value = "/test.html", method = RequestMethod.GET)
 @ResponseBody
 public  User getShopInJSON(User   user ) {

  System.out.println("--------------:"+user.getUsername());
  User shop = new User();
  shop.setId(2);
  shop.setUsername("haha");
  shop.setPassword("zhoushunde");
  shop.setNickname("zxx");
  return shop;

 }

以上以通过;但当用jquery form 提交时就报415错了，

最后换成

public  User add(HttpServletRequest request,
	        HttpServletResponse response ) {
		
		System.out.println("--------------:"+request.getParameter("username"));
		User shop = new User();
		shop.setId(2);
		shop.setUsername("haha");
		shop.setPassword("zhoushunde");
		shop.setNickname("zxx");
		return shop;
}

或者是public  User add(@RequestParam(value = "username", required = true)String  username) {这种也可以通过 public  User add(@RequestParam String  username) {}

form是使用的是post的方式提交，以下对于post方式的处理

var obj={};
		obj.username="zsd";
		var param = JSON.stringify(obj);
		alert(param);
                $.ajax({
		type: "POST",
		contentType: "application/json; charset=utf-8",
		url: "<%=request.getContextPath()%>/test.html",
		data:param,//若这里改成data:{"username":"zsd"}请求会报400 dataType: "json",
		 success: function(text){
		 alert(text.username)
		 },
		 error: function(err) {
			alert("err"); } 
					});

Controller中----------

@RequestMapping(value = "/test.html", method = RequestMethod.POST)
	@ResponseBody
	public  User getShopInJSON(@RequestBody User user) {

		System.out.println("--------------:"+user.getUsername());
		User shop = new User();
		shop.setId(2);
		shop.setUsername("haha");
		shop.setPassword("zhoushunde");
		shop.setNickname("zxx");
		return shop;

	}

转载于:https://my.oschina.net/stonezing/blog/122265