Question

我有一个应该配置的设备。它有一个WEB Interface，IP x.x.x.x，我可以通过IP中的Browser连接到设备并执行它们，现在客户想要一个软件来执行此操作。设备核心是customized Linux。我想找到它通过网络界面发送的Services和Parameters。所以我打开web interface并转到IP和我想改变的部分。在web中更改字段后，我点击了Apply Button。我还下载Fiddler并打开了Developer Tools FireFox。在我点击Developer Tools后的Apply中，会出现一个POST method，其他人都是Get Method。此外，在我点击Apply Button后，实际URL从x.x.x.x/services更改为x.x.x.x/apply.cgi。所以我似乎可以通过x.x.x.x/apply.cgi URL来完成它们。我打开Developer Tools和POST method请求标头就像：

Host: x.x.x.x
User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:51.0)    Gecko/20100101 Firefox/51.0
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
Accept-Language: en-US,en;q=0.5
Accept-Encoding: gzip, deflate
Referer: http://x.x.x.x/Services.asp
Authorization: Basic cm9vdDoxMjM0NQ==
Connection: keep-alive
Upgrade-Insecure-Requests: 1

在Fiddler的TextView我们有这个：

submit_button=Services&action=ApplyTake&change_action=&submit_type=&commit=1&static_leases=13&openvpn_certtype=&dhcpd_usejffs=&dhcpd_usenvram=0
(It is alot, i just copy part of that )
enter code here

我想我可以写一个简单的调用HttpURLConnection通过这种方式，这是我的代码：

    public class PostHTTPSample {
   private String USER_AGENT = "Mozilla/5.0";  
   String userCredentials = "root:xxxxxx";

    public void sendRequestPost() throws IOException { 
    try {
        String encoded = Base64.getEncoder().encodeToString((userCredentials).getBytes(StandardCharsets.UTF_8));  //Java 8
        String str = "http://x.x.x.x/apply.cgi";
        URL obj = new URL(str);
        HttpURLConnection con = (HttpURLConnection) obj.openConnection();
        con.setRequestMethod("POST");
        con.setRequestProperty("User-Agent", USER_AGENT);
        con.setRequestProperty("Accept-Language", "en-US,en;q=0.5");
        con.setRequestProperty("Accept-Encoding", "gzip, deflate");
        con.setRequestProperty("Authorization", "Basic "+"cm9vdDoxMjM0NQ=="); // I even tested encoded String that i has declared and define above
        con.setRequestProperty("Host", "x.x.x.x");
        con.setRequestProperty("Connection", "keep-alive");
        con.setRequestProperty("Upgrade-Insecure-Requests", "1");

        String urlParameters = "The String Copied from Fiddler(Above) For test";
        con.setDoOutput(true);

        DataOutputStream wr = new DataOutputStream(con.getOutputStream());
        wr.writeBytes(urlParameters);
        wr.flush();
        wr.close();
        int responseCode = con.getResponseCode();
        System.out.println("\nSending 'POST' request to URL : " + str);
        System.out.println("Post parameters : " + urlParameters);
        System.out.println("Response Code : " + responseCode);

        BufferedReader in = new BufferedReader(
                new InputStreamReader(con.getInputStream()));
        String inputLine;
        StringBuffer response = new StringBuffer();

        while ((inputLine = in.readLine()) != null) {
            response.append(inputLine);
        }
        in.close();

        //print result
        System.out.println(response.toString());
    } catch (MalformedURLException ex) {
        Logger.getLogger(PostHTTPSample.class.getName()).log(Level.SEVERE, null, ex);
    }

    }

好的，我从这个类中创建一个实例并运行该方法，它给了我Response Code : 400。以及Netbeans的运行控制台：

    Feb 01, 2017 3:16:13 PM posthttpsample.PostHTTPSample main
SEVERE: null
java.io.IOException: Server returned HTTP response code: 400 for URL: http://x.x.x.x/apply.cgi
    at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
    at sun.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:62)
    at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45)
    at java.lang.reflect.Constructor.newInstance(Constructor.java:423)
    at sun.net.www.protocol.http.HttpURLConnection$10.run(HttpURLConnection.java:1926)
    at sun.net.www.protocol.http.HttpURLConnection$10.run(HttpURLConnection.java:1921)
    at java.security.AccessController.doPrivileged(Native Method)
    at sun.net.www.protocol.http.HttpURLConnection.getChainedException(HttpURLConnection.java:1920)
    at sun.net.www.protocol.http.HttpURLConnection.getInputStream0(HttpURLConnection.java:1490)
    at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1474)
    at posthttpsample.PostHTTPSample.sendRequestPost(PostHTTPSample.java:58)
    at posthttpsample.PostHTTPSample.main(PostHTTPSample.java:82)
Caused by: java.io.IOException: Server returned HTTP response code: 400 for URL: http://x.x.x.x/apply.cgi
    at sun.net.www.protocol.http.HttpURLConnection.getInputStream0(HttpURLConnection.java:1876)
    at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1474)
    at java.net.HttpURLConnection.getResponseCode(HttpURLConnection.java:480)
    at posthttpsample.PostHTTPSample.sendRequestPost(PostHTTPSample.java:52)
    ... 1 more
    enter code here

我做错了什么？我该怎么办？有什么建议吗？

编辑：我可以使用此网址的GET方法，它会给我200 OK response。但仍然无法使用POST方法。

通过HttpURLConnection将POST方法发送到未知设备

0 个答案: