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();
}
}
}
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();
}
}
response = requests.get('http://www.yahoo.com/')
;在Java中应该也可以实现类似简洁的代码。 - Dan Passaro我知道其他人会推荐使用Apache的http-client,但它增加了复杂性(即更容易出现错误),而这很少是必要的。对于简单的任务,java.net.URL
足够了。
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();
}
Apache HttpComponents。这两个模块的示例 - HttpCore 和 HttpClient 可以让您立即开始。
并不是说HttpUrlConnection是一个不好的选择,HttpComponents将抽象掉很多繁琐的编码。如果您真的想通过最少的代码支持许多HTTP服务器/客户端,我会推荐使用它。顺便说一下,HttpCore可用于拥有最少功能的应用程序(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支持等功能的客户端。
这里是一个完整的 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://example.com/").openStream())) {
System.out.println(s.useDelimiter("\\A").next());
}
}
}
新的 try-with-resources 会自动关闭 Scanner,进而自动关闭 InputStream。main()
抛出 Exception,其中包括 MalformedURLException 和 IOException。 - jerzy谷歌Java HTTP客户端拥有良好的API,可以进行HTTP请求。你可以轻松地添加JSON支持等等。尽管对于简单的请求来说可能会过于繁琐。
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();
}
}
HTTP_TRANSPORT
,我已经编辑了答案。 - Tombart这会对你有所帮助。别忘了将HttpClient.jar
JAR包添加到类路径中。
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@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();
}
}
}
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();
\r\n
而不是\n
? - CuriousGuy这里有一个非常好的关于发送POST请求的链接,由Example Depot提供:点击这里
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) {
}
wr.write(data);
。conn.setReadTimeout(2000);
,输入参数以毫秒为单位。如果您使用的是Java 11或更新版本(Android除外),则可以使用Java 11新的HTTP Client API,而不是传统的HttpUrlConnection类。
var uri = URI.create("https://httpbin.org/get?age=26&isHappy=true");
var client = HttpClient.newHttpClient();
var request = HttpRequest
.newBuilder()
.uri(uri)
.header("accept", "application/json")
.GET()
.build();
var response = client.send(request, HttpResponse.BodyHandlers.ofString());
System.out.println(response.statusCode());
System.out.println(response.body());
异步执行相同的请求:
var responseAsync = client
.sendAsync(request, HttpResponse.BodyHandlers.ofString())
.thenApply(HttpResponse::body)
.thenAccept(System.out::println);
// responseAsync.join(); // Wait for completion
var request = HttpRequest
.newBuilder()
.uri(uri)
.version(HttpClient.Version.HTTP_2)
.timeout(Duration.ofMinutes(1))
.header("Content-Type", "application/json")
.header("Authorization", "Bearer fake")
.POST(BodyPublishers.ofString("{ title: 'This is cool' }"))
.build();
var response = client.send(request, HttpResponse.BodyHandlers.ofString());
如果要以multipart (multipart/form-data
)或者url-encoded (application/x-www-form-urlencoded
)格式发送表单数据,请参考这个解决方案。
有关HTTP Client API的示例和更多信息,请参见此文章。