如何获取字符串形式的 HTTP 响应正文？

Question

我知道曾经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

...这里有一个示例：

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

...但我相信这是已弃用。

有没有其他方法可以在 Java 中发出 http get 请求并以字符串而不是流的形式获取响应正文？

Answer 1

这是我的工作项目中的两个示例。

1. 使用EntityUtils 和 HttpEntity< /a>

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

2. 使用BasicResponseHandler

   HttpResponse 响应 = 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() 来自 Apache Commons IO 在一个方法调用中将 InputStream 读取到 String 中。例如：

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，以使用响应中的内容编码（如果可用）。否则，它会默认为 UTF-8 作为最佳猜测，而不是使用本地系统默认值。

Answer 3

这是我正在使用 Apache 的 httpclient 库处理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 来获取字符串形式的响应正文。很简单。

Answer 4

这在特定情况下相对简单，但在一般情况下相当棘手。

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 响应标头。

该标头包含有关有效负载的信息，并且可能定义文本数据的编码。即使您假设文本类型，您也可能需要检查内容本身才能确定正确的字符编码。 例如，请参阅 HTML 4 规范 了解详细信息关于如何针对特定格式执行此操作。

一旦编码已知，InputStreamReader 可用于解码数据。

这个答案取决于服务器是否做正确的事情 - 如果您想处理响应标头与文档不匹配的情况，或者文档声明与所使用的编码不匹配的情况，那就是另一回事了。< /em>

Answer 5

下面是使用 Apache HTTP 客户端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

麦克道尔的答案是正确的。但是，如果您尝试上面几篇文章中的其他建议。

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

然后它会给你非法状态异常，指出内容已经被消耗。

Answer 7

就这个怎么样？

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

Answer 8

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 9

我们还可以使用下面的代码来获取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 10

这是一种轻量级的方法：

String responseString = "";
for (int i = 0; i < response.getEntity().getContentLength(); i++) { 
    responseString +=
    Character.toString((char)response.getEntity().getContent().read()); 
}

当然 responseString 包含网站的响应，并且响应是 HttpResponse 类型，由 HttpClient.execute(request)

Answer 11

以下代码片段显示了将响应正文作为字符串处理的更好方法，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 12

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

Answer 13

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();