用 Java 发送 HTTP POST 请求
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Sending HTTP POST Request In Java
提问by Jazz
lets assume this URL...
让我们假设这个网址...
http://www.example.com/page.php?id=10
(Here id needs to be sent in a POST request)
（这里id需要在POST请求中发送）
I want to send the id = 10to the server's page.php, which accepts it in a POST method.
我想将 发送id = 10到服务器的page.php，它以 POST 方法接受它。
How can i do this from within Java?
我如何从 Java 中执行此操作？
I tried this :
我试过这个：
URL aaa = new URL("http://www.example.com/page.php");
URLConnection ccc = aaa.openConnection();
But I still can't figure out how to send it via POST
但我仍然不知道如何通过 POST 发送它
回答by user207421
Call HttpURLConnection.setRequestMethod("POST")and HttpURLConnection.setDoOutput(true);Actually only the latter is needed as POST then becomes the default method.
CallHttpURLConnection.setRequestMethod("POST")并且HttpURLConnection.setDoOutput(true);实际上只需要后者，因为 POST 然后成为默认方法。
回答by DuduAlul
String rawData = "id=10";
String type = "application/x-www-form-urlencoded";
String encodedData = URLEncoder.encode( rawData, "UTF-8" );
URL u = new URL("http://www.example.com/page.php");
HttpURLConnection conn = (HttpURLConnection) u.openConnection();
conn.setDoOutput(true);
conn.setRequestMethod("POST");
conn.setRequestProperty( "Content-Type", type );
conn.setRequestProperty( "Content-Length", String.valueOf(encodedData.length()));
OutputStream os = conn.getOutputStream();
os.write(encodedData.getBytes());
回答by mhshams
Updated Answer:
更新答案：
Since some of the classes, in the original answer, are deprecated in the newer version of Apache HTTP Components, I'm posting this update.
由于原始答案中的某些类在较新版本的 Apache HTTP 组件中已弃用，因此我发布了此更新。
By the way, you can access the full documentation for more examples here.
顺便说一下，您可以在此处访问完整文档以获取更多示例。
HttpClient httpclient = HttpClients.createDefault();
HttpPost httppost = new HttpPost("http://www.a-domain.com/foo/");
// Request parameters and other properties.
List<NameValuePair> params = new ArrayList<NameValuePair>(2);
params.add(new BasicNameValuePair("param-1", "12345"));
params.add(new BasicNameValuePair("param-2", "Hello!"));
httppost.setEntity(new UrlEncodedFormEntity(params, "UTF-8"));
//Execute and get the response.
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
if (entity != null) {
try (InputStream instream = entity.getContent()) {
// do something useful
}
}
Original Answer:
原答案：
I recommend to use Apache HttpClient. its faster and easier to implement.
我建议使用 Apache HttpClient。它更快，更容易实现。
HttpPost post = new HttpPost("http://jakarata.apache.org/");
NameValuePair[] data = {
new NameValuePair("user", "joe"),
new NameValuePair("password", "bloggs")
};
post.setRequestBody(data);
// execute method and handle any error responses.
...
InputStream in = post.getResponseBodyAsStream();
// handle response.
for more information check this url: http://hc.apache.org/
有关更多信息，请查看此网址：http: //hc.apache.org/
回答by Mar Cnu
The first answer was great, but I had to add try/catch to avoid Java compiler errors.
Also, I had troubles to figure how to read the HttpResponsewith Java libraries.
第一个答案很好，但我不得不添加 try/catch 以避免 Java 编译器错误。
另外，我很难弄清楚如何HttpResponse使用 Java 库阅读。
Here is the more complete code :
这是更完整的代码：
/*
* Create the POST request
*/
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://example.com/");
// Request parameters and other properties.
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("user", "Bob"));
try {
httpPost.setEntity(new UrlEncodedFormEntity(params, "UTF-8"));
} catch (UnsupportedEncodingException e) {
// writing error to Log
e.printStackTrace();
}
/*
* Execute the HTTP Request
*/
try {
HttpResponse response = httpClient.execute(httpPost);
HttpEntity respEntity = response.getEntity();
if (respEntity != null) {
// EntityUtils to get the response content
String content = EntityUtils.toString(respEntity);
}
} catch (ClientProtocolException e) {
// writing exception to log
e.printStackTrace();
} catch (IOException e) {
// writing exception to log
e.printStackTrace();
}
回答by Mathias Bak
A simple way using Apache HTTP Components is
使用 Apache HTTP 组件的一种简单方法是
Request.Post("http://www.example.com/page.php")
.bodyForm(Form.form().add("id", "10").build())
.execute()
.returnContent();
Take a look at the Fluent API
回答by Ferrybig
Sending a POST request is easy in vanilla Java. Starting with a URL, we need t convert it to a URLConnectionusing url.openConnection();. After that, we need to cast it to a HttpURLConnection, so we can access its setRequestMethod()method to set our method. We finally say that we are going to send data over the connection.
在 vanilla Java 中发送 POST 请求很容易。从 a 开始URL，我们不需要将其转换为URLConnectionusing url.openConnection();。之后，我们需要将它转换为 a HttpURLConnection，以便我们可以访问它的setRequestMethod()方法来设置我们的方法。我们最后说我们将通过连接发送数据。
URL url = new URL("https://www.example.com/login");
URLConnection con = url.openConnection();
HttpURLConnection http = (HttpURLConnection)con;
http.setRequestMethod("POST"); // PUT is another valid option
http.setDoOutput(true);
We then need to state what we are going to send:
然后我们需要说明我们要发送的内容：
Sending a simple form
发送一个简单的表格
A normal POST coming from a http form has a well definedformat. We need to convert our input to this format:
来自 http 表单的普通 POST 具有明确定义的格式。我们需要将输入转换为这种格式：
Map<String,String> arguments = new HashMap<>();
arguments.put("username", "root");
arguments.put("password", "sjh76HSn!"); // This is a fake password obviously
StringJoiner sj = new StringJoiner("&");
for(Map.Entry<String,String> entry : arguments.entrySet())
sj.add(URLEncoder.encode(entry.getKey(), "UTF-8") + "="
+ URLEncoder.encode(entry.getValue(), "UTF-8"));
byte[] out = sj.toString().getBytes(StandardCharsets.UTF_8);
int length = out.length;
We can then attach our form contents to the http request with proper headers and send it.
然后，我们可以将表单内容附加到带有适当标头的 http 请求并发送。
http.setFixedLengthStreamingMode(length);
http.setRequestProperty("Content-Type", "application/x-www-form-urlencoded; charset=UTF-8");
http.connect();
try(OutputStream os = http.getOutputStream()) {
os.write(out);
}
// Do something with http.getInputStream()
Sending JSON
发送 JSON
We can also send json using java, this is also easy:
我们也可以使用java发送json，这也很简单：
byte[] out = "{\"username\":\"root\",\"password\":\"password\"}" .getBytes(StandardCharsets.UTF_8);
int length = out.length;
http.setFixedLengthStreamingMode(length);
http.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
http.connect();
try(OutputStream os = http.getOutputStream()) {
os.write(out);
}
// Do something with http.getInputStream()
Remember that different servers accept different content-types for json, see thisquestion.
请记住，不同的服务器接受不同的 json 内容类型，请参阅此问题。
Sending files with java post
使用java post发送文件
Sending files can be considered more challenging to handle as the format is more complex. We are also going to add support for sending the files as a string, since we don't want to buffer the file fully into the memory.
由于格式更复杂，因此可以认为发送文件更具挑战性。我们还将添加对将文件作为字符串发送的支持，因为我们不想将文件完全缓冲到内存中。
For this, we define some helper methods:
为此，我们定义了一些辅助方法：
private void sendFile(OutputStream out, String name, InputStream in, String fileName) {
String o = "Content-Disposition: form-data; name=\"" + URLEncoder.encode(name,"UTF-8")
+ "\"; filename=\"" + URLEncoder.encode(filename,"UTF-8") + "\"\r\n\r\n";
out.write(o.getBytes(StandardCharsets.UTF_8));
byte[] buffer = new byte[2048];
for (int n = 0; n >= 0; n = in.read(buffer))
out.write(buffer, 0, n);
out.write("\r\n".getBytes(StandardCharsets.UTF_8));
}
private void sendField(OutputStream out, String name, String field) {
String o = "Content-Disposition: form-data; name=\""
+ URLEncoder.encode(name,"UTF-8") + "\"\r\n\r\n";
out.write(o.getBytes(StandardCharsets.UTF_8));
out.write(URLEncoder.encode(field,"UTF-8").getBytes(StandardCharsets.UTF_8));
out.write("\r\n".getBytes(StandardCharsets.UTF_8));
}
We can then use these methods to create a multipart post request as follows:
然后我们可以使用这些方法来创建一个多部分的 post 请求，如下所示：
String boundary = UUID.randomUUID().toString();
byte[] boundaryBytes =
("--" + boundary + "\r\n").getBytes(StandardCharsets.UTF_8);
byte[] finishBoundaryBytes =
("--" + boundary + "--").getBytes(StandardCharsets.UTF_8);
http.setRequestProperty("Content-Type",
"multipart/form-data; charset=UTF-8; boundary=" + boundary);
// Enable streaming mode with default settings
http.setChunkedStreamingMode(0);
// Send our fields:
try(OutputStream out = http.getOutputStream()) {
// Send our header (thx Algoman)
out.write(boundaryBytes);
// Send our first field
sendField(out, "username", "root");
// Send a seperator
out.write(boundaryBytes);
// Send our second field
sendField(out, "password", "toor");
// Send another seperator
out.write(boundaryBytes);
// Send our file
try(InputStream file = new FileInputStream("test.txt")) {
sendFile(out, "identification", file, "text.txt");
}
// Finish the request
out.write(finishBoundaryBytes);
}
// Do something with http.getInputStream()
回答by chandan kumar
simplest way to send parameters with the post request:
使用 post 请求发送参数的最简单方法：
String postURL = "http://www.example.com/page.php";
HttpPost post = new HttpPost(postURL);
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("id", "10"));
UrlEncodedFormEntity ent = new UrlEncodedFormEntity(params, "UTF-8");
post.setEntity(ent);
HttpClient client = new DefaultHttpClient();
HttpResponse responsePOST = client.execute(post);
You have done. now you can use responsePOST.
Get response content as string:
你已经完成了。现在您可以使用responsePOST. 以字符串形式获取响应内容：
BufferedReader reader = new BufferedReader(new InputStreamReader(responsePOST.getEntity().getContent()), 2048);
if (responsePOST != null) {
StringBuilder sb = new StringBuilder();
String line;
while ((line = reader.readLine()) != null) {
System.out.println(" line : " + line);
sb.append(line);
}
String getResponseString = "";
getResponseString = sb.toString();
//use server output getResponseString as string value.
}
回答by Beno Arakelyan
I recomend use http-requestbuilt on apache http api.
我建议使用基于 apache http api 构建的http-request。
HttpRequest<String> httpRequest = HttpRequestBuilder.createPost("http://www.example.com/page.php", String.class)
.responseDeserializer(ResponseDeserializer.ignorableDeserializer()).build();
public void send(){
String response = httpRequest.execute("id", "10").get();
}
