带有表单数据的 Java Post 请求

Question

我想用 Java 做一个简单的 POST 调用，
我收到了 200 响应代码，但是响应消息错误，
有人告诉我，在使用表单数据时，有一种不同的方式可以进行 Post 调用。

以下是我当前用于进行后期调用的 Java 代码 -

private String makePostCall(){
        try {
            String url = "http://someIp/trusted";
            HttpClient client = new DefaultHttpClient();
            HttpPost post = new HttpPost(url);

            // add header
            List<NameValuePair> urlParameters = new ArrayList<NameValuePair>();
            urlParameters.add(new BasicNameValuePair("username", "app_user"));

            post.setEntity(new UrlEncodedFormEntity(urlParameters));

            HttpResponse response = client.execute(post);
            System.out.println("\nSending 'POST' request to URL : " + url);
            System.out.println("Post parameters : " + post.getEntity());
            System.out.println("Response Code : " + response.getStatusLine().getStatusCode());

            BufferedReader rd = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
            StringBuffer result = new StringBuffer();
            String line = "";
            while ((line = rd.readLine()) != null) {
                result.append(line);
            }

            System.out.println(result.toString());
            return result.toString();

        } catch (Exception e) {
            e.printStackTrace();
            return null;
        }
    }

以下是通过 Postman 应用程序运行的 Post 调用示例 -

我指的是以下网站 -
https://www.mkyong.com/java/how-to-send-http-request-getpost-in-java/

后调用的预期结果应该是一个令牌，即。 字符串值，当前响应为 -1。

Answer 1

这些答案是正确的，但我很难把它变成一个工作代码。 因此，让我提供一个可以重复使用的通用且有效的答案。

private static RequestConfig requestConfig = RequestConfig.custom().build();

public HttpResponse postWithFormData(String url, List<NameValuePair> params) throws IOException {
        // building http client
        HttpClient httpClient = HttpClientBuilder.create().setDefaultRequestConfig(requestConfig).build();
        HttpPost request = new HttpPost(url);

        // adding the form data
        request.setEntity(new UrlEncodedFormEntity(params));
        return httpClient.execute(request);
    }

List<NameValuePair> urlParameters = new ArrayList<>();

// add any number of form data
urlParameters.add(new BasicNameValuePair("form_key_1", "form_value_1");
urlParameters.add(new BasicNameValuePair("form_key_2", "form_value_2");

// Getting the HTTP Response and processing it
HttpResponse response = postWithFormData("http_url", urlParameters);
HttpEntity entity = response.getEntity();
// String of the response
String responseString = EntityUtils.toString(entity);
// JSON of the response (use this only if the response is a JSON)
JSONObject responseObject = new JSONObject(responseString);

这些是我的主要导入，以防有人对导入的内容感到困惑。

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.message.BasicNameValuePair;
import org.apache.http.util.EntityUtils;
import org.json.JSONObject;
import org.apache.http.client.HttpClient;
import org.apache.http.client.entity.UrlEncodedFormEntity;

Answer 2

通过显式设置内容类型multipart/form-data尝试一下，

post.setHeader("Content-Type", "multipart/form-data");

在您的代码中，

post.setEntity(new UrlEncodedFormEntity(urlParameters)); 
post.setHeader("Content-Type", "multipart/form-data");