在Java中,如何撰写HTTP请求消息并将其发送到HTTP web服务器?
您可以使用java.net.HttpUrlConnection。
示例(来自这里),并进行了改进。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();
}
}
}
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();
}
}
我知道其他人会推荐Apache的http-客户端,但是它增加了复杂性(例如,更多可能出错的东西),这是很少被保证的。对于简单的任务,可以使用java.net.URL。
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。这两个模块的例子——HttpCore和HttpClient会让你马上开始。
并不是说HttpUrlConnection是一个糟糕的选择,HttpComponents将抽象出大量繁琐的编码。如果你真的想用最少的代码来支持大量的HTTP服务器/客户端,我推荐这样做。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用于需要支持多种身份验证方案、cookie支持等的客户端。
这里有一个通过Example Depot发送POST请求的链接::
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) {
}
如果您想发送一个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,还注释掉这个wr.write(data);
有一件事没有写下来,你应该注意,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,或者至少等待很长时间,这可能是你不想要的。
超时设置如下conn.setReadTimeout(2000);输入参数以毫秒为单位
这对你有帮助。不要忘记将JAR HttpClient.jar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();
}
}
}
你可以使用Socket来实现
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();
下面是一个完整的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将自动关闭Scanner,而Scanner将自动关闭InputStream。
谷歌java http客户端有不错的API http请求。您可以轻松地添加JSON支持等。虽然对于简单的要求来说可能有点过分。
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();
}
}
如果您正在使用Java 11或更新版本(Android除外),而不是遗留的HttpUrlConnection类,您可以使用Java 11新的HTTP客户端API。
一个示例GET请求:
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());
同一请求异步执行:
var responseAsync = client
.sendAsync(request, HttpResponse.BodyHandlers.ofString())
.thenApply(HttpResponse::body)
.thenAccept(System.out::println);
// responseAsync.join(); // Wait for completion
一个POST请求的例子:
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());
要以多部分(multipart/form-data)或url编码(application/x-www-form-urlencoded)格式发送表单数据,请参阅此解决方案。
有关HTTP客户端API的示例和更多信息,请参阅本文。
对于Java标准库HTTP服务器,请参阅这篇文章。
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