ajax post提交多个参数后台controller @RequestParam方式接收

注意： 返回数据类型用dataType: ‘text’,如果是 dataType: ‘json’,success获取不到

####1.ajax post提交多个参数后台controller @RequestParam方式接收

@RequestMapping(value = “/findPage”,method = RequestMethod.POST)
public String doFindPageByPageNum(@RequestParam(value = “page”) String page, @RequestParam String size,@RequestParam String cartonDetailNum,String cartoonId){
//代码省略
}

这里写代码片

####例子2：

controller接收：

 @RequestMapping(value = "/findVideoByCategoryId",method = RequestMethod.POST)
    @ResponseBody
    public String findVideoByCategoryId( String pageIndex,
                                         String pageSize,  String categoryId,HttpServletRequest request,HttpServletResponse response){
                                         //logger.info("----ajax POST 方式提交后台接收-----");
                                         //HttpServletRequest，HttpServletResponse只是用于验证登录token
     }

####2. jquery ajax get方式提交单个参数后台controller @RequestParam方式接收

确定

controller:

@RequestMapping(value = "/doWxPayTest",method = RequestMethod.GET)
    @ResponseBody
    public String dowxPayTest(Model model,@RequestParam String totalFee){ 
        if(!totalFee.equals("")&&totalFee!=null) {
            total_fee = Integer.parseInt(totalFee);
        }
        logger.info("wxPayTest total_fee="+total_fee);
        return "y";
    }

最后附上，忘记时查看
后台接收POST GET 提交参数的方式总结