我正在使用这个函数-
public BufferedReader GetResponse(String url, String urlParameters){
HttpsURLConnection con=null;
try{
URL obj = new URL(url);
con = (HttpsURLConnection) obj.openConnection();
//add reuqest header
con.setRequestMethod("POST");
// Send post request
con.setDoOutput(true);
DataOutputStream wr = new DataOutputStream(con.getOutputStream());
wr.writeBytes(urlParameters);
wr.flush();
wr.close();
int responseCode = con.getResponseCode();
BufferedReader returnValue= new BufferedReader(new InputStreamReader(con.getInputStream()));
con.disconnect();
return returnValue;
}
catch(Exception e){
System.out.println("--------------------There is an error in response function ");
e.printStackTrace();
LOGGER.error("There is an arror in AjaxCAll function of login controller."+e.getMessage());
LOGGER.debug("There is an arror in AjaxCAll function of login controller.");
return null;
}
finally{
}
}
如果我使用这个 -
con.disconnect();
然后我得到java.io.IOException:流是关闭错误,但是如果我评论 con.disconnect() 行,那么一切正常。我不知道为什么会这样。
调用函数
BufferedReader rd = utilities.GetResponse(url, urlParameters);
// Send post request
String line = "";
try{
while ((line = rd.readLine()) != null) { //getting error here if i close connection in response method
// Parse our JSON response
responsString += line;
JSONParser j = new JSONParser();
JSONObject o = (JSONObject) j.parse(line);
if (o.containsKey("response")) {
restMessage = (Map) o.get("response");
} else {
restMessage = (Map) o;
}
}
} finally{
rd.close();
}
来自 JavaDoc of HttpURLConnection
(HttpsURLConnection 扩展):
调用 disconnect() 方法可能会关闭底层套接字,如果此时持久连接处于空闲状态。
在你的GetResponse()
方法中,你得到了一个引用HttpsURLConnection
的InputStream
作为BufferedReader
。 但是,当您使用 con.disconnect()
时,您关闭了该基础InputStream
。
在调用 GetResponse()
方法的代码中,当您稍后尝试使用返回的 BufferedReader
时,您会得到一个java.io.IOException: stream is closed error
,因为您已经使用 con.disconnect()
间接关闭了该流。
您需要重新排列代码,以便在完成BufferedReader
之前不调用con.disconnect()
。
这是一种方法:
GetResponse():
public HttpsURLConnection GetResponse(String url, String urlParameters) {
HttpsURLConnection con = null;
DataOutputStream wr = null;
try{
URL obj = new URL(url);
con = (HttpsURLConnection) obj.openConnection();
//add request header
con.setRequestMethod("POST");
// Send post request
con.setDoOutput(true);
wr = new DataOutputStream(con.getOutputStream());
wr.writeBytes(urlParameters);
wr.flush();
}
catch(Exception e) { //better to catch more specific than Exception
System.out.println("--------------------There is an error in response function ");
e.printStackTrace();
LOGGER.error("There is an arror in AjaxCAll function of login controller."+e.getMessage());
LOGGER.debug("There is an arror in AjaxCAll function of login controller.");
return null;
}
finally{
if(wr != null) {
wr.close();
}
}
return con;
}
电话代码:
HttpsURLConnection con = utilities.GetResponse(url, urlParameters);
if(con == null) {
//stop processing or maybe throw an exception depending on what you want to do.
}
BufferedReader rd = null;
// Send post request
String line = "";
try{
int responseCode = con.getResponseCode(); //what's this for? nothing is being done with the variable
rd = new BufferedReader(new InputStreamReader(con.getInputStream()));
while ((line = rd.readLine()) != null) {
// Parse our JSON response
responsString += line;
JSONParser j = new JSONParser();
JSONObject o = (JSONObject) j.parse(line);
if (o.containsKey("response")) {
restMessage = (Map) o.get("response");
}
else {
restMessage = (Map) o;
}
}
}
catch(Exception e) { //better to catch more specific than Exception
//handle exception
}
finally {
if(rd != null) {
rd.close();
}
con.disconnect(); //we already checked if null above
}