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  • Spring MVC 的@RequestParam注解和request.getParameter("XXX")

    在SpringMVC后台控制层获取参数的方式主要有两种,一种是request.getParameter("name"),另外一种是用注解@RequestParam直接获取。这里主要讲这个注解 

    一、基本使用,获取提交的参数 
    后端代码:
     

    Java代码  收藏代码
    1. @RequestMapping("testRequestParam")    
    2.    public String filesUpload(@RequestParam String inputStr, HttpServletRequest request) {    
    3.     System.out.println(inputStr);  
    4.       
    5.     int inputInt = Integer.valueOf(request.getParameter("inputInt"));  
    6.     System.out.println(inputInt);  
    7.       
    8.     // ......省略  
    9.     return "index";  
    10.    }     



    前端代码: 

    Html代码  收藏代码
    1. <form action="/gadget/testRequestParam" method="post">    
    2.      参数inputStr:<input type="text" name="inputStr">    
    3.      参数intputInt:<input type="text" name="inputInt">    
    4. </form>  



    前端界面: 
     

    执行结果: 
    test1 
    123 

    可以看到spring会自动根据参数名字封装进入,我们可以直接拿这个参数名来用 

    二、各种异常情况处理 
    1、可以对传入参数指定参数名
     

    Java代码  收藏代码
    1. @RequestParam String inputStr  
    2. // 下面的对传入参数指定为aa,如果前端不传aa参数名,会报错  
    3. @RequestParam(value="aa") String inputStr  


    错误信息: 
    HTTP Status 400 - Required String parameter 'aa' is not present 

    2、可以通过required=false或者true来要求@RequestParam配置的前端参数是否一定要传
     

    Java代码  收藏代码
    1. // required=false表示不传的话,会给参数赋值为null,required=true就是必须要有  
    2. @RequestMapping("testRequestParam")    
    3.     public String filesUpload(@RequestParam(value="aa", required=true) String inputStr, HttpServletRequest request)  



    3、如果用@RequestMapping注解的参数是int基本类型,但是required=false,这时如果不传参数值会报错,因为不传值,会赋值为null给int,这个不可以 

    Java代码  收藏代码
    1. @RequestMapping("testRequestParam")    
    2.    public String filesUpload(@RequestParam(value="aa", required=true) String inputStr,   
    3.         @RequestParam(value="inputInt", required=false) int inputInt  
    4.         ,HttpServletRequest request) {    
    5.       
    6.     // ......省略  
    7.     return "index";  
    8.    }  


    解决方法: 
        “Consider declaring it as object wrapper for the corresponding primitive type.”建议使用包装类型代替基本类型,如使用“Integer”代替“int”
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  • 原文地址:https://www.cnblogs.com/baobeiqi-e/p/9884741.html
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