Question

我知道过去有一种方法可以通过apache commons获取它，如下所示： http://hc.apache.org/httpclient-legacy/apidocs/org/apache/commons/httpclient/HttpMethod.html 这里有一个例子：

http://www.kodejava.org/examples/416.html

但我相信这已被弃用。是否还有其他方法可以在java中生成http get请求并将响应主体作为字符串而不是流来获取？

Answer 1

以下是我工作项目的两个例子。

使用EntityUtils和HttpEntity

HttpResponse response = httpClient.execute(new HttpGet(URL));
HttpEntity entity = response.getEntity();
String responseString = EntityUtils.toString(entity, "UTF-8");
System.out.println(responseString);

使用BasicResponseHandler

HttpResponse response = httpClient.execute(new HttpGet(URL));
String responseString = new BasicResponseHandler().handleResponse(response);
System.out.println(responseString);

Answer 2

我能想到的每个库都会返回一个流。您可以使用IOUtils.toString()中的Apache Commons IO在一次方法调用中将InputStream读入String。 E.g：

URL url = new URL("http://www.example.com/");
URLConnection con = url.openConnection();
InputStream in = con.getInputStream();
String encoding = con.getContentEncoding();
encoding = encoding == null ? "UTF-8" : encoding;
String body = IOUtils.toString(in, encoding);
System.out.println(body);

更新：我更改了上面的示例，以便在响应中使用内容编码（如果可用）。否则它将默认为UTF-8作为最佳猜测，而不是使用本地系统默认值。

Answer 3

以下是我正在使用Apache的httpclient库的另一个简单项目的示例：

String response = new String();
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(1);
nameValuePairs.add(new BasicNameValuePair("j", request));
HttpEntity requestEntity = new UrlEncodedFormEntity(nameValuePairs);

HttpPost httpPost = new HttpPost(mURI);
httpPost.setEntity(requestEntity);
HttpResponse httpResponse = mHttpClient.execute(httpPost);
HttpEntity responseEntity = httpResponse.getEntity();
if(responseEntity!=null) {
    response = EntityUtils.toString(responseEntity);
}

只需使用EntityUtils将响应主体作为String抓取。非常简单。

Answer 4

在特定情况下这是相对简单的，但在一般情况下非常棘手。

HttpClient httpclient = new DefaultHttpClient();
HttpGet httpget = new HttpGet("http://stackoverflow.com/");
HttpResponse response = httpclient.execute(httpget);
HttpEntity entity = response.getEntity();
System.out.println(EntityUtils.getContentMimeType(entity));
System.out.println(EntityUtils.getContentCharSet(entity));

答案取决于Content-Type HTTP response header。

此标头包含有关有效负载的信息，可能定义文本数据的编码。即使您假设text types，也可能需要检查内容本身以确定正确的字符编码。 E.g。有关如何为该特定格式执行此操作的详细信息，请参阅HTML 4 spec。

知道编码后，可以使用InputStreamReader对数据进行解码。

这个答案取决于服务器做正确的事情 - 如果你想处理响应头与文档不匹配的情况，或者文档声明与使用的编码不匹配，那就是另一个水壶鱼。

Answer 5

下面是使用Apache HTTP Client库以字符串形式访问响应的简单方法。

import org.apache.http.HttpResponse;
import org.apache.http.client.HttpClient;
import org.apache.http.client.ResponseHandler;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.BasicResponseHandler;

//... 

HttpGet get;
HttpClient httpClient;

// initialize variables above

ResponseHandler<String> responseHandler = new BasicResponseHandler();
String responseBody = httpClient.execute(get, responseHandler);

Answer 6

这个怎么样？

org.apache.commons.io.IOUtils.toString(new URL("http://www.someurl.com/"));

Answer 7

McDowell的回答是正确的。但是，如果你在上面几篇文章中尝试其他建议。

HttpEntity responseEntity = httpResponse.getEntity();
if(responseEntity!=null) {
   response = EntityUtils.toString(responseEntity);
   S.O.P (response);
}

然后它将为您提供illegalStateException，声明内容已被使用。

Answer 8

我们也可以使用以下代码来获取java中的HTML响应

import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.HttpResponse;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import org.apache.log4j.Logger;

public static void main(String[] args) throws Exception {
    HttpClient client = new DefaultHttpClient();
    //  args[0] :-  http://hostname:8080/abc/xyz/CheckResponse
    HttpGet request1 = new HttpGet(args[0]);
    HttpResponse response1 = client.execute(request1);
    int code = response1.getStatusLine().getStatusCode();

    try (BufferedReader br = new BufferedReader(new InputStreamReader((response1.getEntity().getContent())));) {
        // Read in all of the post results into a String.
        String output = "";
        Boolean keepGoing = true;
        while (keepGoing) {
            String currentLine = br.readLine();

            if (currentLine == null) {
                keepGoing = false;
            } else {
                output += currentLine;
            }
        }

        System.out.println("Response-->" + output);
    } catch (Exception e) {
        System.out.println("Exception" + e);

    }
}

Answer 9

这是一种轻量级的方法：

objectB

当然String responseString = ""; for (int i = 0; i < response.getEntity().getContentLength(); i++) { responseString += Character.toString((char)response.getEntity().getContent().read()); }包含网站的回复，回复的类型为responseString，由HttpResponse

返回

Answer 10

以下是代码段，该段显示了将响应正文作为String处理的更好方法，无论它是HTTP POST请求的有效响应还是错误响应：

BufferedReader reader = null;
OutputStream os = null;
String payload = "";
try {
    URL url1 = new URL("YOUR_URL");
    HttpURLConnection postConnection = (HttpURLConnection) url1.openConnection();
    postConnection.setRequestMethod("POST");
    postConnection.setRequestProperty("Content-Type", "application/json");
    postConnection.setDoOutput(true);
    os = postConnection.getOutputStream();
    os.write(eventContext.getMessage().getPayloadAsString().getBytes());
    os.flush();

    String line;
    try{
        reader = new BufferedReader(new InputStreamReader(postConnection.getInputStream()));
    }
    catch(IOException e){
        if(reader == null)
            reader = new BufferedReader(new InputStreamReader(postConnection.getErrorStream()));
    }
    while ((line = reader.readLine()) != null)
        payload += line.toString();
}       
catch (Exception ex) {
            log.error("Post request Failed with message: " + ex.getMessage(), ex);
} finally {
    try {
        reader.close();
        os.close();
    } catch (IOException e) {
        log.error(e.getMessage(), e);
        return null;
    }
}

Answer 11

您可以使用发送Http请求并处理响应的3-d参与者库。其中一种著名的产品是Apache commons HTTPClient：HttpClient javadoc，HttpClient Maven artifact。到目前为止，鲜为人知但更简单的HTTPClient（由我编写的开源MgntUtils库的一部分）：MgntUtils HttpClient javadoc，MgntUtils maven artifact，MgntUtils Github。使用这些库中的任何一个，您都可以独立于Spring发送REST请求并接收响应，作为业务逻辑的一部分

Answer 12

如果您正在使用Jackson来反序列化响应正文，则一个非常简单的解决方案是使用request.getResponseBodyAsStream()而不是 ~~request.getResponseBodyAsString()~~

Answer 13

这是一个普通的 Java 答案：

import java.net.http.HttpClient;
import java.net.http.HttpResponse;
import java.net.http.HttpRequest;
import java.net.http.HttpRequest.BodyPublishers;

...
HttpClient client = HttpClient.newHttpClient();
HttpRequest request = HttpRequest.newBuilder()
  .uri(targetUrl)
  .header("Content-Type", "application/json")
  .POST(BodyPublishers.ofString(requestBody))
  .build();
HttpResponse response = client.send(request, HttpResponse.BodyHandlers.ofString());
String responseString = (String) response.body();

Answer 14

使用Apache commons Fluent API，可以按如下方式完成，

String response = Request.Post("http://www.example.com/")
                .body(new StringEntity(strbody))
                .addHeader("Accept","application/json")
                .addHeader("Content-Type","application/json")
                .execute().returnContent().asString();

如何在Java中将http响应主体作为字符串获取？

14 个答案: