我成功地使用这段代码通过GET方法发送带有一些参数的HTTP请求

void sendRequest(String request)
{
    // i.e.: request = "http://example.com/index.php?param1=a&param2=b&param3=c";
    URL url = new URL(request); 
    HttpURLConnection connection = (HttpURLConnection) url.openConnection();           
    connection.setDoOutput(true); 
    connection.setInstanceFollowRedirects(false); 
    connection.setRequestMethod("GET"); 
    connection.setRequestProperty("Content-Type", "text/plain"); 
    connection.setRequestProperty("charset", "utf-8");
    connection.connect();
}

现在我可能需要通过POST方法发送参数(即param1, param2, param3),因为它们非常长。 我想添加一个额外的参数,该方法(即字符串httpMethod)。

我怎样才能尽可能少地更改上面的代码,以便能够通过GET或POST发送参数?

我希望情况会有所改变

connection.setRequestMethod("GET");

to

connection.setRequestMethod("POST");

可以完成这个任务,但是参数仍然通过GET方法发送。

HttpURLConnection有任何方法可以帮助吗? 有什么有用的Java构造吗?

任何帮助都将不胜感激。


当前回答

在GET请求中,参数作为URL的一部分发送。

在POST请求中,参数作为请求体发送,位于请求头之后。

要使用HttpURLConnection进行POST,需要在打开连接后将参数写入连接。

这段代码应该让你开始:

String urlParameters  = "param1=a&param2=b&param3=c";
byte[] postData       = urlParameters.getBytes( StandardCharsets.UTF_8 );
int    postDataLength = postData.length;
String request        = "http://example.com/index.php";
URL    url            = new URL( request );
HttpURLConnection conn= (HttpURLConnection) url.openConnection();           
conn.setDoOutput( true );
conn.setInstanceFollowRedirects( false );
conn.setRequestMethod( "POST" );
conn.setRequestProperty( "Content-Type", "application/x-www-form-urlencoded"); 
conn.setRequestProperty( "charset", "utf-8");
conn.setRequestProperty( "Content-Length", Integer.toString( postDataLength ));
conn.setUseCaches( false );
try( DataOutputStream wr = new DataOutputStream( conn.getOutputStream())) {
   wr.write( postData );
}

其他回答

在GET请求中,参数作为URL的一部分发送。

在POST请求中,参数作为请求体发送,位于请求头之后。

要使用HttpURLConnection进行POST,需要在打开连接后将参数写入连接。

这段代码应该让你开始:

String urlParameters  = "param1=a&param2=b&param3=c";
byte[] postData       = urlParameters.getBytes( StandardCharsets.UTF_8 );
int    postDataLength = postData.length;
String request        = "http://example.com/index.php";
URL    url            = new URL( request );
HttpURLConnection conn= (HttpURLConnection) url.openConnection();           
conn.setDoOutput( true );
conn.setInstanceFollowRedirects( false );
conn.setRequestMethod( "POST" );
conn.setRequestProperty( "Content-Type", "application/x-www-form-urlencoded"); 
conn.setRequestProperty( "charset", "utf-8");
conn.setRequestProperty( "Content-Length", Integer.toString( postDataLength ));
conn.setUseCaches( false );
try( DataOutputStream wr = new DataOutputStream( conn.getOutputStream())) {
   wr.write( postData );
}

我看到一些其他的答案给出了另一种选择,我个人认为直觉上你在做正确的事情;)不好意思,在devoxx上,有几个演讲者一直在抱怨这种事情。

这就是为什么我个人使用Apache的HTTPClient/HttpCore库来做这类工作,我发现它们的API比Java的原生HTTP支持更容易使用。当然是YMMV了!

GET和POST方法设置如下…两种类型的api调用1)get()和2)post()。Get()方法从API json数组中获取值& post()方法在url中的数据post中使用并获得响应。

 public class HttpClientForExample {

    private final String USER_AGENT = "Mozilla/5.0";

    public static void main(String[] args) throws Exception {

        HttpClientExample http = new HttpClientExample();

        System.out.println("Testing 1 - Send Http GET request");
        http.sendGet();

        System.out.println("\nTesting 2 - Send Http POST request");
        http.sendPost();

    }

    // HTTP GET request
    private void sendGet() throws Exception {

        String url = "http://www.google.com/search?q=developer";

        HttpClient client = new DefaultHttpClient();
        HttpGet request = new HttpGet(url);

        // add request header
        request.addHeader("User-Agent", USER_AGENT);

        HttpResponse response = client.execute(request);

        System.out.println("\nSending 'GET' request to URL : " + url);
        System.out.println("Response Code : " + 
                       response.getStatusLine().getStatusCode());

        BufferedReader rd = new BufferedReader(
                       new InputStreamReader(response.getEntity().getContent()));

        StringBuffer result = new StringBuffer();
        String line = "";
        while ((line = rd.readLine()) != null) {
            result.append(line);
        }

        System.out.println(result.toString());

    }

    // HTTP POST request
    private void sendPost() throws Exception {

        String url = "https://selfsolve.apple.com/wcResults.do";

        HttpClient client = new DefaultHttpClient();
        HttpPost post = new HttpPost(url);

        // add header
        post.setHeader("User-Agent", USER_AGENT);

        List<NameValuePair> urlParameters = new ArrayList<NameValuePair>();
        urlParameters.add(new BasicNameValuePair("sn", "C02G8416DRJM"));
        urlParameters.add(new BasicNameValuePair("cn", ""));
        urlParameters.add(new BasicNameValuePair("locale", ""));
        urlParameters.add(new BasicNameValuePair("caller", ""));
        urlParameters.add(new BasicNameValuePair("num", "12345"));

        post.setEntity(new UrlEncodedFormEntity(urlParameters));

        HttpResponse response = client.execute(post);
        System.out.println("\nSending 'POST' request to URL : " + url);
        System.out.println("Post parameters : " + post.getEntity());
        System.out.println("Response Code : " + 
                                    response.getStatusLine().getStatusCode());

        BufferedReader rd = new BufferedReader(
                        new InputStreamReader(response.getEntity().getContent()));

        StringBuffer result = new StringBuffer();
        String line = "";
        while ((line = rd.readLine()) != null) {
            result.append(line);
        }

        System.out.println(result.toString());

    }

}

我也有同样的问题。我想通过POST发送数据。 我使用了以下代码:

    URL url = new URL("http://example.com/getval.php");
    Map<String,Object> params = new LinkedHashMap<>();
    params.put("param1", param1);
    params.put("param2", param2);

    StringBuilder postData = new StringBuilder();
    for (Map.Entry<String,Object> param : params.entrySet()) {
        if (postData.length() != 0) postData.append('&');
        postData.append(URLEncoder.encode(param.getKey(), "UTF-8"));
        postData.append('=');
        postData.append(URLEncoder.encode(String.valueOf(param.getValue()), "UTF-8"));
    }
    String urlParameters = postData.toString();
    URLConnection conn = url.openConnection();

    conn.setDoOutput(true);

    OutputStreamWriter writer = new OutputStreamWriter(conn.getOutputStream());

    writer.write(urlParameters);
    writer.flush();

    String result = "";
    String line;
    BufferedReader reader = new BufferedReader(new InputStreamReader(conn.getInputStream()));

    while ((line = reader.readLine()) != null) {
        result += line;
    }
    writer.close();
    reader.close()
    System.out.println(result);

我使用Jsoup进行解析:

    Document doc = Jsoup.parseBodyFragment(value);
    Iterator<Element> opts = doc.select("option").iterator();
    for (;opts.hasNext();) {
        Element item = opts.next();
        if (item.hasAttr("value")) {
            System.out.println(item.attr("value"));
        }
    }

似乎你还必须调用connection. getoutputstream ()"至少一次"(以及setDoOutput(true)),以将其视为POST。

所以最低要求的代码是:

    URL url = new URL(urlString);
    HttpURLConnection connection = (HttpURLConnection) url.openConnection();
    //connection.setRequestMethod("POST"); this doesn't seem to do anything at all..so not useful
    connection.setDoOutput(true); // set it to POST...not enough by itself however, also need the getOutputStream call...
    connection.connect();
    connection.getOutputStream().close(); 

令人惊讶的是,你甚至可以在urlString中使用“GET”样式参数。尽管这可能会让事情变得复杂。

显然,你也可以使用NameValuePair。