如何在java中发送HTTP请求？
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How to send HTTP request in java?
提问by Yatendra Goel
In Java, How to compose a HTTP request message and send it to a HTTP WebServer?
在 Java 中，如何编写 HTTP 请求消息并将其发送到 HTTP WebServer？
采纳答案by duffymo
You can use java.net.HttpUrlConnection.
您可以使用java.net.HttpUrlConnection。
Example (from here), with improvements. Included in case of link rot:
示例（来自此处），进行了改进。包括在链接腐烂的情况下：
public static String executePost(String targetURL, String urlParameters) {
HttpURLConnection connection = null;
try {
//Create connection
URL url = new URL(targetURL);
connection = (HttpURLConnection) url.openConnection();
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type",
"application/x-www-form-urlencoded");
connection.setRequestProperty("Content-Length",
Integer.toString(urlParameters.getBytes().length));
connection.setRequestProperty("Content-Language", "en-US");
connection.setUseCaches(false);
connection.setDoOutput(true);
//Send request
DataOutputStream wr = new DataOutputStream (
connection.getOutputStream());
wr.writeBytes(urlParameters);
wr.close();
//Get Response
InputStream is = connection.getInputStream();
BufferedReader rd = new BufferedReader(new InputStreamReader(is));
StringBuilder response = new StringBuilder(); // or StringBuffer if Java version 5+
String line;
while ((line = rd.readLine()) != null) {
response.append(line);
response.append('\r');
}
rd.close();
return response.toString();
} catch (Exception e) {
e.printStackTrace();
return null;
} finally {
if (connection != null) {
connection.disconnect();
}
}
}
回答by Chi
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();
}
}
回答by erickson
I know others will recommend Apache's http-client, but it adds complexity (i.e., more things that can go wrong) that is rarely warranted. For a simple task, java.net.URLwill do.
我知道其他人会推荐 Apache 的 http 客户端，但它增加了复杂性（即，更多可能出错的事情），这是很少有保证的。对于一个简单的任务，java.net.URL会做。
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();
}
回答by Vineet Reynolds
Apache HttpComponents. The examples for the two modules - HttpCoreand HttpClientwill get you started right away.
Apache HttpComponents。两个模块的示例 - HttpCore和HttpClient将让您立即开始。
Not that HttpUrlConnection is a bad choice, HttpComponents will abstract a lot of the tedious coding away. I would recommend this, if you really want to support a lot of HTTP servers/clients with minimum code. By the way, HttpCore could be used for applications (clients or servers) with minimum functionality, whereas HttpClient is to be used for clients that require support for multiple authentication schemes, cookie support etc.
并不是说 HttpUrlConnection 是一个糟糕的选择，HttpComponents 会抽象掉很多繁琐的编码。如果你真的想用最少的代码支持很多 HTTP 服务器/客户端，我会推荐这个。顺便说一下，HttpCore 可用于功能最少的应用程序（客户端或服务器），而 HttpClient 可用于需要支持多种身份验证方案、cookie 支持等的客户端。
回答by tzik
There's a great link about sending a POST request hereby Example Depot::
Example Depot有一个关于在这里发送 POST 请求的很好的链接：
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) {
}
If you want to send a GET request you can modify the code slightly to suit your needs. Specifically you have to add the parameters inside the constructor of the URL. Then, also comment out this wr.write(data);
如果您想发送 GET 请求，您可以稍微修改代码以满足您的需要。具体来说，您必须在 URL 的构造函数中添加参数。然后，也注释掉这个wr.write(data);
One thing that's not written and you should beware of, is the timeouts. Especially if you want to use it in WebServices you have to set timeouts, otherwise the above code will wait indefinitely or for a very long time at least and it's something presumably you don't want.
一件事没有写出来，你应该小心，就是超时。特别是如果你想在 WebServices 中使用它，你必须设置超时，否则上面的代码将无限期地等待或至少等待很长时间，这可能是你不想要的。
Timeouts are set like this conn.setReadTimeout(2000);the input parameter is in milliseconds
超时设置如下conn.setReadTimeout(2000);输入参数以毫秒为单位
回答by Satish Sharma
This will help you. Don't forget to add the JAR HttpClient.jarto the classpath.
这会帮助你。不要忘记将 JAR 添加HttpClient.jar到类路径中。
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 protected]");
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();
}
}
}
回答by laksys
You may use Socket for this like
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();
回答by Janus Troelsen
Here's a complete Java 7 program:
这是一个完整的 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());
}
}
}
The new try-with-resources will auto-close the Scanner, which will auto-close the InputStream.
新的 try-with-resources 将自动关闭 Scanner，后者将自动关闭 InputStream。
回答by Tombart
Google java http clienthas nice API for http requests. You can easily add JSON support etc. Although for simple request it might be overkill.
谷歌java http 客户端有很好的 http 请求 API。您可以轻松添加 JSON 支持等。虽然对于简单的请求，它可能有点矫枉过正。
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();
}
}
