如何在Java中发送HTTP请求?\[重复\]

Java HTML

飞云

2020-03-23

在Java中,如何编写HTTP请求消息并将其发送到HTTP WebServer?

第2644篇《如何在Java中发送HTTP请求?\[重复\]》来自Winter(https://github.com/aiyld/aiyld.github.io)的站点

8个回答
Green猿古一 2020.03.23

有一个关于发送POST请求有很大的联系在这里通过实例车厂::

try {
    // Construct data
    String data = URLEncoder.encode("key1", "UTF-8") + "=" + URLEncoder.encode("value1", "UTF-8");
    data += "&" + URLEncoder.encode("key2", "UTF-8") + "=" + URLEncoder.encode("value2", "UTF-8");

    // Send data
    URL url = new URL("http://hostname:80/cgi");
    URLConnection conn = url.openConnection();
    conn.setDoOutput(true);
    OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
    wr.write(data);
    wr.flush();

    // Get the response
    BufferedReader rd = new BufferedReader(new InputStreamReader(conn.getInputStream()));
    String line;
    while ((line = rd.readLine()) != null) {
        // Process line...
    }
    wr.close();
    rd.close();
} catch (Exception e) {
}

如果要发送GET请求,则可以略微修改代码以适合您的需求。具体来说,您必须在URL的构造函数中添加参数。然后,也将其注释掉wr.write(data);

没有写的一件事是您应该注意的,它是超时。特别是如果要在WebServices中使用它,则必须设置超时,否则上面的代码将无限期地等待,或者至少等待很长时间,这大概是您不想要的。

这样设置超时时间conn.setReadTimeout(2000);,输入参数以毫秒为单位

神乐 2020.03.23

您可以像这样使用Socket

String host = "www.yourhost.com";
Socket socket = new Socket(host, 80);
String request = "GET / HTTP/1.0\r\n\r\n";
OutputStream os = socket.getOutputStream();
os.write(request.getBytes());
os.flush();

InputStream is = socket.getInputStream();
int ch;
while( (ch=is.read())!= -1)
    System.out.print((char)ch);
socket.close();    
番长樱梅 2020.03.23

Google Java http客户端对HTTP请求具有不错的API。您可以轻松地添加JSON支持等。尽管对于简单的请求而言,这可能会过分杀伤。

import com.google.api.client.http.GenericUrl;
import com.google.api.client.http.HttpRequest;
import com.google.api.client.http.HttpResponse;
import com.google.api.client.http.HttpTransport;
import com.google.api.client.http.javanet.NetHttpTransport;
import java.io.IOException;
import java.io.InputStream;

public class Network {

    static final HttpTransport HTTP_TRANSPORT = new NetHttpTransport();

    public void getRequest(String reqUrl) throws IOException {
        GenericUrl url = new GenericUrl(reqUrl);
        HttpRequest request = HTTP_TRANSPORT.createRequestFactory().buildGetRequest(url);
        HttpResponse response = request.execute();
        System.out.println(response.getStatusCode());

        InputStream is = response.getContent();
        int ch;
        while ((ch = is.read()) != -1) {
            System.out.print((char) ch);
        }
        response.disconnect();
    }
}
斯丁前端 2020.03.23

这将为您提供帮助。不要忘记将JAR添加HttpClient.jar到类路径。

import java.io.FileOutputStream;
import java.io.IOException;

import org.apache.commons.httpclient.HttpClient;
import org.apache.commons.httpclient.HttpStatus;
import org.apache.commons.httpclient.NameValuePair;
import org.apache.commons.httpclient.methods.PostMethod;

public class MainSendRequest {

     static String url =
         "http://localhost:8080/HttpRequestSample/RequestSend.jsp";

    public static void main(String[] args) {

        //Instantiate an HttpClient
        HttpClient client = new HttpClient();

        //Instantiate a GET HTTP method
        PostMethod method = new PostMethod(url);
        method.setRequestHeader("Content-type",
                "text/xml; charset=ISO-8859-1");

        //Define name-value pairs to set into the QueryString
        NameValuePair nvp1= new NameValuePair("firstName","fname");
        NameValuePair nvp2= new NameValuePair("lastName","lname");
        NameValuePair nvp3= new NameValuePair("email","email@email.com");

        method.setQueryString(new NameValuePair[]{nvp1,nvp2,nvp3});

        try{
            int statusCode = client.executeMethod(method);

            System.out.println("Status Code = "+statusCode);
            System.out.println("QueryString>>> "+method.getQueryString());
            System.out.println("Status Text>>>"
                  +HttpStatus.getStatusText(statusCode));

            //Get data as a String
            System.out.println(method.getResponseBodyAsString());

            //OR as a byte array
            byte [] res  = method.getResponseBody();

            //write to file
            FileOutputStream fos= new FileOutputStream("donepage.html");
            fos.write(res);

            //release connection
            method.releaseConnection();
        }
        catch(IOException e) {
            e.printStackTrace();
        }
    }
}
凯西里 2020.03.23

Apache HttpComponents这两个模块的示例-HttpCoreHttpClient将立即使您入门。

并不是说HttpUrlConnection是一个不好的选择,HttpComponents会将许多繁琐的编码抽象掉。如果您确实想以最少的代码支持许多HTTP服务器/客户端,我会建议您这样做。顺便说一下,HttpCore可以用于功能最少的应用程序(客户端或服务器),而HttpClient则用于需要支持多种身份验证方案,cookie支持等的客户端。

老丝阿飞 2020.03.23

我知道其他人会推荐Apache的http-client,但是它增加了很少需要的复杂性(例如,更多可能出错的东西)。对于一个简单的任务,java.net.URL将做。

URL url = new URL("http://www.y.com/url");
InputStream is = url.openStream();
try {
  /* Now read the retrieved document from the stream. */
  ...
} finally {
  is.close();
}
宝儿 2020.03.23

这是完整的Java 7程序:

class GETHTTPResource {
  public static void main(String[] args) throws Exception {
    try (java.util.Scanner s = new java.util.Scanner(new java.net.URL("http://tools.ietf.org/rfc/rfc768.txt").openStream())) {
      System.out.println(s.useDelimiter("\\A").next());
    }
  }
}

新的try-with-resources将自动关闭Scanner,这将自动关闭InputStream。

老丝阿飞 2020.03.23

来自Oracle的Java教程

import java.net.*;
import java.io.*;

public class URLConnectionReader {
    public static void main(String[] args) throws Exception {
        URL yahoo = new URL("http://www.yahoo.com/");
        URLConnection yc = yahoo.openConnection();
        BufferedReader in = new BufferedReader(
                                new InputStreamReader(
                                yc.getInputStream()));
        String inputLine;

        while ((inputLine = in.readLine()) != null) 
            System.out.println(inputLine);
        in.close();
    }
}

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