RESTAPI可以从Postman调用，显示来自Java的400个错误请求

Question

我正在尝试使用POST-restful服务。当我在邮递员身上试用时，我得到了成功的回复。但当我用下面的代码在java上尝试时，我得到的响应代码是400。在《邮递员》中，我用替换转义字符粘贴相同的输入

try {

    URL url = new URL("https://localhost/PasswordVault/api/Accounts");
    HttpURLConnection conn = (HttpURLConnection) url.openConnection();
    conn.setRequestMethod("POST");
    conn.setDoOutput(true);

    conn.setRequestProperty("Content-Type", "application/json");
    String input = "{\"qty\":100,\"name\":\"iPad 4\", \"detail\":{\"ver\":\"2020\",\"productionDate\":\"2020\"}}";

    OutputStream os = conn.getOutputStream();
    os.write(input.getBytes());
    os.flush();


    if (conn.getResponseCode() != HttpURLConnection.HTTP_CREATED) {
        throw new RuntimeException("Failed : HTTP error code : "
                + conn.getResponseCode());
    }

    BufferedReader br = new BufferedReader(new InputStreamReader(conn.getInputStream()));

    String output;
    System.out.println("Output from Server .... \n");
    while ((output = br.readLine()) != null) {
        System.out.println(output);
    }

    conn.disconnect();
} catch (Exception e) {
        System.out.println("Exception:- " + e);
}

Answer 1

首先，检查邮递员可能正在发送的请求头，而您的代码没有。在发送请求之前，您可能需要在请求中添加一些标题。其次，我强烈建议使用一些3d party Http客户端库，而不是自己编写Http请求。一些著名的库是Apache Http client和OK Http client。我个人使用自己的Http客户端，它是MgntUtils开源库的一部分。你也可以试试。以下是简化的代码，仅用于演示如何从网站上读取一些响应：

    try {
        HttpClient client = new HttpClient();
        client.setConnectionUrl("https://www.yahoo.com/");
        String result = client.sendHttpRequest(HttpMethod.GET);
        System.out.println(result);
    } catch (IOException e) {
        System.out.println(TextUtils.getStacktrace(e, "com.mgnt."));
    }
    

当然，您也可以使用这个库发送POST请求、设置请求头、读取文本或二进制响应和响应头。这里是HttpClient class Javadoc的链接。该库本身以Maven artifact和Github的形式提供，包括Javadoc和源代码

Answer 2

我强烈建议你使用图书馆来做这件事。 您可以使用Jackson，这将极大地简化和标准化java与远程HTTP服务的交互

以下几点会让你开始：

import javax.ws.rs.client.Client;
import javax.ws.rs.client.ClientBuilder;
import javax.ws.rs.client.Entity;
import javax.ws.rs.client.WebTarget;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Response;

public class ApiClient
{

    private transient Client client;
    protected transient WebTarget apiRoot;

    public ApiClient(String rootUrl) 
    {
        this.client = ClientBuilder.newBuilder().build();
        // The root URL is your API starting point, e.g. https://host/api
        this.apiRoot = client.target(rootUrl);
    }

    public Response doPostToAccounts(String data)
    {
        try 
        {
            // This is where you execute your POST request
            return apiRoot.path("/Accounts")
                .request(MediaType.APPLICATION_JSON)
                .post(Entity.entity(data, MediaType.APPLICATION_JSON));
        } catch (Exception e) {
            throw new IllegalStateException(e);
        }
    }

}

Answer 3

一些想法：

• 用UTF-8编码你的请求帖子数据："post data".getBytes(StandardCharsets.UTF_8)
• 阅读服务器的响应（不仅仅是代码），希望能获得有关问题的更多详细信息（connection.getErrorStream()）
• 是否缺少授权标头？（API令牌等）
• 使用单引号使请求数据字符串更具可读性，或者使用JSON库避免手动错误
• 也可以调用OutputStream。（（）冲洗后关闭）

例如

  // handle error code
  if(connection.getResponseCode() != HttpURLConnection.HTTP_OK) {
    // use getErrorStream for non OK statuses (https://stackoverflow.com/questions/613307/read-error-response-body-in-java)
    String errorMessage = "HTTP response code: " + responseCode
        + (responseMessage == null || responseMessage.trim().length() == 0 ? "" : " " + responseMessage);

    InputStream errorStream = connection.getErrorStream();
    if(errorStream != null) {
      String errorString = streamToString(connection.getErrorStream());
      if(errorString.length() > MAX_MSG_LENGTH) {
        errorString = errorString.substring(0, Math.min(MAX_MSG_LENGTH, errorString.length())) + "...";
      }
      errorMessage += ", HTTP response data: " + errorString;
    }

    throw new RuntimeException(errorMessage);
  }

  // handle OK code
  // ...