Question

这是我的代码，我试图将用户详细信息作为json输入传递，但我无法通过rest api方法接收数据。我将所有值都设为null

这是我的json请求

{
    "userId" : "12345",
    "username" : "arun.ammasai",
    "createdBy" : "-2",
    "updatedBy" : "-2",
    "statusCd" : "New",
    "createdDate" : "2019-03-03",
    "updatedDate" : "2019-03-03"
}

================================================ =====================

@RequestMapping(value = "/registerUser", method = RequestMethod.POST, consumes = { "application/JSON", "application/XML" })
private String registerUser(User user) {
    System.out.println(user.toString());
    return "User Created";
}

================================================ =====================

这是Postman Client中的回复

意外的“ U”

Answer 1

使用@RequestBody批注更新您的方法签名。它将自动将json反序列化为Java实体。请注意，json中的名称应与User对象中的参数相同，并且User对象应具有getter和setter。因此您的方法应类似于

    @RequestMapping(value = "/registerUser", method = RequestMethod.POST, consumes = { "application/JSON", "application/XML" })
        private String registerUser(@RequestBody User user) {
            System.out.println(user.toString()); //What is the reason of doing toString of java Object? 
            //better to do System.out.println(user.getUsername())
            return "User Created";
    }

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1 个答案: