当我在浏览器中点击以下网址时,它运行正常:
http://62.215.226.164/fccsms_P.aspx?UID=something&P=something&S=InfoText&G=96567771404&M=hello&L=E
但是当我尝试使用以下java代码点击URL时,它无法正常工作:
try {
URL url = new URL(null, "http://62.215.226.164/fccsms_P.aspx?", new sun.net.www.protocol.https.Handler());
HttpsURLConnection con = (HttpsURLConnection) url.openConnection();
con.setRequestMethod("POST");
String urlParameters="UID=something&P=something=InfoText&G=96567771404&M=hello&L=E";
// Send post request
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 : " + url);
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();
}
catch(Exception e) {
System.err.println("Error: "+e.getMessage());
}
代码中有什么问题吗?
答案 0 :(得分:0)
您正在通过HttpsURLConnection发送HTTP网址。所以它的制作问题。只需更改前2行。你会得到正确的输出。
URL url = new URL("http://62.215.226.164/fccsms_P.aspx");
HttpURLConnection con = (HttpURLConnection) url.openConnection();
答案 1 :(得分:0)
您需要先修复您的请求参数:
urlParameters = “UID =东西&P=something=InfoText
&安培; G = 96567771404&安培; M =&你好放大器; L = E”;
请在此处发布异常堆栈跟踪以帮助您更好。