HTTP 请求 400错误-Toy模板网

这篇具有很好参考价值的文章主要介绍了HTTP 请求 400错误。希望对大家有所帮助。如果存在错误或未考虑完全的地方，请大家不吝赐教，您也可以点击"举报违法"按钮提交疑问。

问题

HTTP 请求 400错误

详细问题

客户端发送请求

public static UserInfo updateUserInfo(UserInfo userInfo) {
    // 创建 OkHttpClient 对象
    OkHttpClient client = new OkHttpClient();
    // 创建请求体
    MediaType JSON = MediaType.parse("application/json; charset=utf-8");
    RequestBody requestBody = RequestBody.create(JSON,
        "{\"id\": " + userInfo.getId() +
        ", \"avatarBase64\"" + userInfo.getAvatarBase64() + "\"" +
        ", \"nickname\": \"" + userInfo.getNickname() + "\"" +
        ", \"bio\": \"" + userInfo.getBio() + "\"" +
        ", \"birthday\": \"" + userInfo.getBirthday() + "\"}");
    // 创建请求对象
    Request request = new Request.Builder()
        .url("http://192.168.43.127:9090/user_info/update_user_info_by_id")
        .post(requestBody)
        .build();

    // 发送请求并处理响应
    client.newCall(request).enqueue(new Callback() {@
        Override
        public void onResponse(Call call, Response response) throws IOException {
            // 处理响应，这里根据具体情况处理更新成功或失败的逻辑
            String responseData = response.body().string();
            // 在这里处理服务器返回的数据
            System.out.println("responseData" + responseData);

        }

        @
        Override
        public void onFailure(Call call, IOException e) {
            // 处理请求失败的情况
            e.printStackTrace();
        }
    });
    return userInfo;
}

服务端接收请求

@PostMapping("/update_user_info_by_id")
public int updateUserInfo(@RequestBody UserInfo userInfo) {
    return userInfoService.updateUserInfo(userInfo);
}

客户端控制台打印

2024-02-17 13:45:21.584 27938-28019/com.example.assistingagriculture I/System.out: responseData{"timestamp":"2024-02-17T05:45:22.846+00:00","status":400,"error":"Bad Request","path":"/user_info/update_user_info_by_id"}

服务端控制台打印

2024-02-17 13:45:22.841  WARN 26516 --- [nio-9090-exec-6] .w.s.m.s.DefaultHandlerExceptionResolver : Resolved [org.springframework.http.converter.HttpMessageNotReadableException: JSON parse error: Unexpected character ('1' (code 49)): was expecting a colon to separate field name and value; nested exception is com.fasterxml.jackson.core.JsonParseException: Unexpected character ('1' (code 49)): was expecting a colon to separate field name and value<EOL> at [Source: (PushbackInputStream); line: 1, column: 27]]

解决方案

将\"avatarBase64\""修改为\"avatarBase64\": \""
即修改请求体：

RequestBody requestBody = RequestBody.create(JSON,
    "{\"id\": " + userInfo.getId() +
    ", \"avatarBase64\"" + userInfo.getAvatarBase64() + "\"" +
    ", \"nickname\": \"" + userInfo.getNickname() + "\"" +
    ", \"bio\": \"" + userInfo.getBio() + "\"" +
    ", \"birthday\": \"" + userInfo.getBirthday() + "\"}");

为

RequestBody requestBody = RequestBody.create(JSON,
    "{\"id\": " + userInfo.getId() +
    ", \"avatarBase64\": \"" + userInfo.getAvatarBase64() + "\"" +
    ", \"nickname\": \"" + userInfo.getNickname() + "\"" +
    ", \"bio\": \"" + userInfo.getBio() + "\"" +
    ", \"birthday\": \"" + userInfo.getBirthday() + "\"}");