在Java中,如何撰写HTTP请求消息并将其发送到HTTP web服务器?
当前回答
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();
}
}
其他回答
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();
}
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。
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();
}
}
}
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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