无法从opend url获取inputstream？

我想从我的web应用程序使用短信网关发送短信。在以下代码中，con.getInputStream();在有控制到达时不起作用，程序抛出异常。无法从opend url获取inputstream？

public String process_sms(String mob_no,String message) throws IOException, KeyManagementException, NoSuchAlgorithmException 
    {  
     message=URLEncoder.encode(message, "UTF-8");     
     URL url = new URL("https://instantalerts.co/api/web/send/?apikey=6d6ra0u305nggr0cvrxxxxxxxxxxxxxx&sender=xxxxxx&to=xxxxxxxxxx&message=Your One Time Password is {$No} "); 
     System.out.println("url look like " + url); 
     HttpURLConnection con = (HttpURLConnection) url.openConnection(); 
     System.out.println("url opend" ); 
     con.setRequestMethod("GET"); 
     System.out.println("url method" ); 
     con.setDoOutput(true); 
     System.out.println("url output" ); 
     con.getOutputStream(); 
     System.out.println("url ouotput2" ); 
     con.getInputStream(); 
     System.out.println("url input" ); 
     BufferedReader rd; 
     String line; 
     String result = ""; 
     rd = new BufferedReader(new InputStreamReader(con.getInputStream())); 
     System.out.println("url input reader" ); 
     while ((line = rd.readLine()) != null) 
     { 
      System.out.println("url input line" ); 
      result += line; 
     } 
     rd.close(); 
     System.out.println("Result is" + result); 
     return result;    
    }

在控制台它打印，直到url ouotput2后con.getInputStream();不工作。我不知道什么是问题。任何人都可以帮我解决这个问题。

错误：

type Exception report 

message Server returned HTTP response code: 403 for URL: https://instantalerts.co/api/web/send/?apikey=6d6ra0u305nggr0cvrxxxxxxxxxxxxxx&sender=xxxxx&to=xxxxxxxxx&message=Your One Time Password is {$No} 

description The server encountered an internal error that prevented it from fulfilling this request. 

exception 

java.io.IOException: Server returned HTTP response code: 403 for URL: https://instantalerts.co/api/web/send/?apikey=6d6ra0u305nggr0cvrxxxxxxxxxxxxx&sender=xxxxxx&to=xxxxxxxxxxx&message=Your One Time Password is {$No} 
    sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1628) 
    sun.net.www.protocol.https.HttpsURLConnectionImpl.getInputStream(HttpsURLConnectionImpl.java:254) 
    send_sms.process_sms(send_sms.java:92) 
    send_sms.doPost(send_sms.java:58) 
    javax.servlet.http.HttpServlet.service(HttpServlet.java:650) 
    javax.servlet.http.HttpServlet.service(HttpServlet.java:731) 
    org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52) 

note The full stack trace of the root cause is available in the Apache Tomcat/7.0.62 logs.

我url.but在我的计划中提到的“apikey”“XXXXXXXXXXXXXXX”，“发件人”和“到”的参数我使用过它是由网关给出供应商。

来源

2016-04-29 KVK

HTTP返回码403表示“禁止”。我想你的网址或你的api键是错误的。 –

@ThomasStets不，我只是复制并通过浏览器选项卡中的网址，它工作正常。 – KVK

@KVK 403意味着禁止。您在使用apikey或网址时出现问题。可能的检查标题以及某些cookie。 – user2494817

为什么你会得到OutputStream呢？然后不发送什么？这只适用于请求方法POST。

同样，你打开输入流两次 - 第二个流应该从哪里来？

试试这样说：

public String process_sms(String mob_no,String message) throws IOException, KeyManagementException, NoSuchAlgorithmException 
{  
    message=URLEncoder.encode(message, "UTF-8");     
    URL url = new URL("https://instantalerts.co/api/web/send/?apikey=6d6ra0u305nggr0cvrxxxxxxxxxxxxxx&sender=xxxxxx&to=xxxxxxxxxx&message=Your One Time Password is {$No} "); 
    HttpURLConnection con = (HttpURLConnection) url.openConnection(); 
    System.out.println("url opend" ); 
    String line; 
    String result = ""; 
    BufferedReader rd = new BufferedReader(new InputStreamReader(con.getInputStream())); 
    System.out.println("url input reader: " + rd); 
    while ((line = rd.readLine()) != null) 
    { 
     System.out.println("url input line" ); 
     result += line; 
    } 
    rd.close(); 
    System.out.println("Result is" + result); 
    return result;    
}

来源

2016-04-29 06:51:26 Jan

我累了，但它显示我无效模板error.because我使用短信网关的帐户帐户。 – KVK

那么现在你确实从SMS网关获取了一些东西？所以与* InputStream *问题似乎解决了？ – Jan

您的API URL被认为只能做到单词“你”如下：

https://instantalerts.co/api/web/send/?apikey=6d6ra0u305nggr0cvrxxxxxxxxxxxxxx&sender=xxxxxx&to=xxxxxxxxxx&message=Your

正因如此它与响应消息字之间的空间403错误

所以，你必须使用URLEncoder对网址进行编码：

String message = URLEncoder.encode("Your One Time Password is {$No}", "UTF-8"); 
URL url = new URL("https://instantalerts.co/api/web/send/?apikey=6d6ra0u305nggr0cvrxxxxxxxxxxxxxx&sender=xxxxxx&to=xxxxxxxxxx&message=" + message);

来源

2017-05-05 13:14:50

无法从opend url获取inputstream？

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