From: rishi p. <mai...@gm...> - 2006-06-28 12:30:00
|
---------- Forwarded message ---------- From: rishi pathak <mai...@gm...> Date: Jun 28, 2006 5:34 PM Subject: Digital(certificate) signing of Applet -still gives the problem To: jyt...@li..., jyt...@li... Hi all I built a simple applet in jython which on keyTyped event sends the typed character to the server(written in python listining on 80 port). The archive for the applet is a jar file But when I try to view it on a browser , it gives me the following error's java.security.AccessControlException: access denied ( java.net.SocketPermission 192.168.2.40:80 connect,resolve) at java.security.AccessControlContext.checkPermission( AccessControlContext.java:264) at java.security.AccessController.checkPermission( AccessController.java:427) at java.lang.SecurityManager.checkPermission(SecurityManager.java :532) at java.lang.SecurityManager.checkConnect(SecurityManager.java:1034) at java.net.Socket.connect(Socket.java:501) at java.net.Socket.connect(Socket.java:457) at java.net.Socket.<init>(Socket.java:365) at java.net.Socket.<init>(Socket.java:178) at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method) at sun.reflect.NativeConstructorAccessorImpl.newInstance( NativeConstructorAccessorImpl.java:39) at sun.reflect.DelegatingConstructorAccessorImpl.newInstance( DelegatingConstructorAccessorImpl.java:27) at java.lang.reflect.Constructor.newInstance(Constructor.java:494) at org.python.core.PyReflectedConstructor.__call__( PyReflectedConstructor.java) The code for the applet is: from java.applet import Applet from java.awt.event import KeyListener from java.awt import * from java.net import Socket from java.io import * class shell1(Applet,KeyListener): key='' def init(self): self.addKeyListener(self) self.socket=csSocket() def keyTyped(self,keyEvent): self.key=keyEvent.getKeyChar() print self.key self.socket.csWrite(self.key) self.repaint() def paint(self,g): g.setColor(Color.black) g.setColor(Color.white) g.drawString(str(self.key),50,50) def keyReleased(self,keyEvent): pass def keyPressed(self,keyEvent): pass def stop(self): self.cs.close() class csSocket(Socket): HOST='192.168.2.40' PORT=80 def __init__(self): self.cs=Socket(self.HOST,self.PORT) self.outputStream=cs.getOutputStream() self.inputStream=cs.getInputStream() def csRead(self): self.inputStream.read() def csWrite(self,data): self.outputStream.write(data) def csClose(self): self.cs.close() The server is running on 192.168.2.40:80 and the applet on 192.168.2.31 I compiled the .py file with following syntax jythonc --jar shell1.jar -a -d -J "-source 1.2" shell1.py Then on the local machine where the applet resides I generated a key with: keytool -genkey //I did'nt gave the key name and so it created key with keyname as 'mykey' Then I fired the following commands: keytool -selfcert jarsigner shell1.jar mykey It exited successfully // That means shell1.jar got signed(am I right or wrong) But still when I view the applet it repeats the above mentioned error's Do some one have the solution for this Would be grateful if I get some - Regards - Rishi Pathak |
From: rishi p. <mai...@gm...> - 2006-06-28 13:46:25
|
Hi all I built a simple applet in jython which on keyTyped event sends the typed character to the server(written in python listining on 80 port). The archive for the applet is a jar file But when I try to view it on a browser , it gives me the following error's java.security.AccessControlException: access denied ( java.net.SocketPermission 192.168.2.40:80 connect,resolve) at java.security.AccessControlContext.checkPermission( AccessControlContext.java:264) at java.security.AccessController.checkPermission( AccessController.java:427) at java.lang.SecurityManager.checkPermission(SecurityManager.java :532) at java.lang.SecurityManager.checkConnect(SecurityManager.java:1034) at java.net.Socket.connect(Socket.java:501) at java.net.Socket.connect(Socket.java:457) at java.net.Socket.<init>(Socket.java:365) at java.net.Socket.<init>(Socket.java:178) at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method) at sun.reflect.NativeConstructorAccessorImpl.newInstance( NativeConstructorAccessorImpl.java:39) at sun.reflect.DelegatingConstructorAccessorImpl.newInstance( DelegatingConstructorAccessorImpl.java:27) at java.lang.reflect.Constructor.newInstance(Constructor.java:494) at org.python.core.PyReflectedConstructor.__call__( PyReflectedConstructor.java) The code for the applet is: from java.applet import Applet from java.awt.event import KeyListener from java.awt import * from java.net import Socket from java.io import * class shell1(Applet,KeyListener): key='' def init(self): self.addKeyListener(self) self.socket=csSocket() def keyTyped(self,keyEvent): self.key=keyEvent.getKeyChar() print self.key self.socket.csWrite(self.key) self.repaint() def paint(self,g): g.setColor(Color.black) g.setColor(Color.white) g.drawString(str(self.key),50,50) def keyReleased(self,keyEvent): pass def keyPressed(self,keyEvent): pass def stop(self): self.cs.close() class csSocket(Socket): HOST='192.168.2.40' PORT=80 def __init__(self): self.cs=Socket(self.HOST,self.PORT) self.outputStream=cs.getOutputStream() self.inputStream=cs.getInputStream() def csRead(self): self.inputStream.read() def csWrite(self,data): self.outputStream.write(data) def csClose(self): self.cs.close() The server is running on 192.168.2.40:80 and the applet on 192.168.2.31 I compiled the .py file with following syntax jythonc --jar shell1.jar -a -d -J "-source 1.2" shell1.py Then on the local machine where the applet resides I generated a key with: keytool -genkey //I did'nt gave the key name and so it created key with keyname as 'mykey' Then I fired the following commands: keytool -selfcert jarsigner shell1.jar mykey It exited successfully // That means shell1.jar got signed(am I right or wrong) But still when I view the applet it repeats the above mentioned error's Do some one have the solution for this Would be grateful if I get some - Regards - Rishi Pathak |
From: rishi p. <mai...@gm...> - 2006-06-28 14:19:34
|
Hi all I built a simple applet in jython which on keyTyped event sends the typed character to the server(written in python listining on 80 port). The archive for the applet is a jar file But when I try to view it on a browser , it gives me the following error's java.security.AccessControlExce ption: access denied ( java.net.SocketPermission 192.168.2.40:80 <http://192.168.2.40/>connect,resolve) at java.security.AccessControlContext.checkPermission( AccessControlContext.java:264) at java.security.AccessController.checkPermission( AccessController.java:427) at java.lang.SecurityManager.checkPermission(SecurityManager.java :532) at java.lang.SecurityManager.checkConnect(SecurityManager.java:1034) at java.net.Socket.connect(Socket.java:501) at java.net.Socket.connect(Socket.java:457) at java.net.Socket.<init>(Socket.java:365) at java.net.Socket.<init>(Socket.java:178) at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method) at sun.reflect.NativeConstructorAccessorImpl.newInstance( NativeConstructorAccessorImpl.java:39) at sun.reflect.DelegatingConstructorAccessorImpl.newInstance( DelegatingConstructorAccessorImpl.java:27) at java.lang.reflect.Constructor.newInstance(Constructor.java:494) at org.python.core.PyReflectedConstructor.__call__( PyReflectedConstructor.java) The code for the applet is: from java.applet import Applet from java.awt.event import KeyListener from java.awt import * from java.net import Socket from java.io import * class shell1(Applet,KeyListener): key='' def init(self): self.addKeyListener(self) self.socket=csSocket() def keyTyped(self,keyEvent): self.key=keyEvent.getKeyChar() print self.key self.socket.csWrite(self.key) self.repaint() def paint(self,g): g.setColor(Color.black) g.setColor(Color.white) g.drawString(str(self.key),50,50) def keyReleased(self,keyEvent): pass def keyPressed(self,keyEvent): pass def stop(self): self.cs.close() class csSocket(Socket): HOST='192.168.2.40' PORT=80 def __init__(self): self.cs=Socket(self.HOST,self.PORT) self.outputStream=cs.getOutputStream() self.inputStream=cs.getInputStream() def csRead(self): self.inputStream.read() def csWrite(self,data): self.outputStream.write(data) def csClose(self): self.cs.close() The server is running on 192.168.2.40:80 <http://192.168.2.40/> and the applet on 192.168.2.31 I compiled the .py file with following syntax jythonc --jar shell1.jar -a -d -J "-source 1.2" shell1.py Then on the local machine where the applet resides I generated a key with: keytool -genkey //I did'nt gave the key name and so it created key with keyname as 'mykey' Then I fired the following commands: keytool -selfcert jarsigner shell1.jar mykey It exited successfully // That means shell1.jar got signed(am I right or wrong) But still when I view the applet it repeats the above mentioned error's Do some one have the solution for this Would be grateful if I get some - Regards |
From: Badal, C. F <chr...@hp...> - 2006-06-28 14:43:04
|
With the exception shown below, it certainly appears as though the applet was not signed. Did you get a popup that asks you if you wish to run the applet? (If not, the applet was not signed properly) =20 Use the following link for a step by step guide to signing an applet. =20 http://forum.java.sun.com/thread.jspa?threadID=3D174214&start=3D0&tstart=3D= 0 =20 -Chris =20 =20 ________________________________ From: jyt...@li... [mailto:jyt...@li...] On Behalf Of rishi pathak Sent: Wednesday, June 28, 2006 6:32 AM To: jyt...@li... Subject: [Jython-users] Digital(certificate) signing of Applet -still givesthe problem Hi all I built a simple applet in jython which on keyTyped event sends the typed character to the server(written in python listining on 80 port). The archive for the applet is a jar file But when I try to view it on a browser , it gives me the following error's java.security.AccessControlExce=20 ption: access denied (java.net.SocketPermission 192.168.2.40:80 <http://192.168.2.40/> connect,resolve) at java.security.AccessControlContext.checkPermission(AccessControlContext. java:264) at java.security.AccessController.checkPermission(AccessController.java:427 ) at java.lang.SecurityManager.checkPermission(SecurityManager.java:532) at java.lang.SecurityManager.checkConnect(SecurityManager.java:1034) at java.net.Socket.connect(Socket.java:501) at java.net.Socket.connect(Socket.java:457) at java.net.Socket.<init>(Socket.java:365) at java.net.Socket.<init>(Socket.java:178) at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method) at sun.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorA ccessorImpl.java:39) at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingCons tructorAccessorImpl.java:27) at java.lang.reflect.Constructor.newInstance(Constructor.java:494) at org.python.core.PyReflectedConstructor.__call__(PyReflectedConstructor.j ava) The code for the applet is: from java.applet import Applet from java.awt.event import KeyListener from java.awt import * from java.net <http://java.net/> import Socket from java.io <http://java.io/> import * class shell1(Applet,KeyListener): key=3D'' def init(self): self.addKeyListener(self) self.socket=3DcsSocket() def keyTyped(self,keyEvent): self.key=3DkeyEvent.getKeyChar() print self.key self.socket.csWrite(self.key) self.repaint() def paint(self,g): g.setColor(Color.black) g.setColor(Color.white) g.drawString(str(self.key),50,50) def keyReleased(self,keyEvent): pass def keyPressed(self,keyEvent): pass def stop(self): self.cs.close() class csSocket(Socket): HOST=3D'192.168.2.40 <http://192.168.2.40/> ' PORT=3D80 def __init__(self): self.cs=3DSocket(self.HOST,self.PORT) self.outputStream=3Dcs.getOutputStream() self.inputStream=3Dcs.getInputStream() def csRead(self): self.inputStream.read() def csWrite(self,data): self.outputStream.write(data) def csClose(self): self.cs.close() The server is running on 192.168.2.40:80 <http://192.168.2.40/> and the applet on 192.168.2.31 <http://192.168.2.31/>=20 I compiled the .py file with following syntax jythonc --jar shell1.jar -a -d -J "-source 1.2" shell1.py Then on the local machine where the applet resides I generated a key with: keytool -genkey //I did'nt gave the key name and so it created key with keyname as 'mykey' Then I fired the following commands: keytool -selfcert jarsigner shell1.jar mykey It exited successfully // That means shell1.jar got signed(am I right or wrong) But still when I view the applet it repeats the above mentioned error's Do some one have the solution for this=20 Would be grateful if I get some - Regards |
From: Dong, M. <MDong@StateStreet.com> - 2006-06-28 16:11:10
|
Hi, Usually the exception message is printed out on the stdout such as Traceback (innermost last): File "./ACC/acclog.py", line 15, in ? NameError: existingProps=20 Can the exception message be retrieved in jython code as a string? It will be greatly appreciated if you can give me a hand.=20 Thanks a lot. Ming Dong |
From: Jeff E. <jem...@fr...> - 2006-06-28 17:02:45
|
sys.exc_info returns a tuple containing the exception class, exception instance, and a traceback object. You can generate a string like that printed from the exc info. import sys ei=None try: doSomethingBad() except: ei=sys.exc_info() Another option would be to implement an excepthook function that replaces stderr before calling the original excepthook. The replaced stderr would capture the stream output to a string. Dong, Ming wrote: > > Hi, > > Usually the exception message is printed out on the stdout such as > Traceback (innermost last): > File "./ACC/acclog.py", line 15, in ? > NameError: existingProps > > Can the exception message be retrieved in jython code as a string? It > will be greatly appreciated if you can give me a hand. > > Thanks a lot. > > Ming Dong > > Using Tomcat but need to do more? Need to support web services, security? > Get stuff done quickly with pre-integrated technology to make your job easier > Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo > http://sel.as-us.falkag.net/sel?cmd=lnk&kid=120709&bid=263057&dat=121642 > _______________________________________________ > Jython-users mailing list > Jyt...@li... > https://lists.sourceforge.net/lists/listinfo/jython-users |
From: Raghuram D. <ra...@sy...> - 2006-06-28 17:29:33
|
I use a function like: def get_excinfo_str(): (exc_type, exc_value, exc_traceback) = sys.exc_info() formatted_excinfo = traceback.format_exception(exc_type, exc_value, exc_traceback) excinfo_str = string.join(formatted_excinfo) del exc_type del exc_value del exc_traceback return(excinfo_str) and call this from exception handling code. Raghu. Jeff Emanuel wrote: > sys.exc_info returns a tuple containing the > exception class, exception instance, and > a traceback object. You can generate > a string like that printed from the exc > info. > > import sys > ei=None > try: > doSomethingBad() > except: > ei=sys.exc_info() > > > Another option would be to implement an > excepthook function that replaces stderr > before calling the original excepthook. > The replaced stderr would capture the > stream output to a string. > > > > Dong, Ming wrote: >> Hi, >> >> Usually the exception message is printed out on the stdout such as >> Traceback (innermost last): >> File "./ACC/acclog.py", line 15, in ? >> NameError: existingProps >> >> Can the exception message be retrieved in jython code as a string? It >> will be greatly appreciated if you can give me a hand. >> >> Thanks a lot. >> >> Ming Dong >> >> Using Tomcat but need to do more? Need to support web services, security? >> Get stuff done quickly with pre-integrated technology to make your job easier >> Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo >> http://sel.as-us.falkag.net/sel?cmd=lnk&kid=120709&bid=263057&dat=121642 >> _______________________________________________ >> Jython-users mailing list >> Jyt...@li... >> https://lists.sourceforge.net/lists/listinfo/jython-users > > Using Tomcat but need to do more? Need to support web services, security? > Get stuff done quickly with pre-integrated technology to make your job easier > Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo > http://sel.as-us.falkag.net/sel?cmd=lnk&kid=120709&bid=263057&dat=121642 > _______________________________________________ > Jython-users mailing list > Jyt...@li... > https://lists.sourceforge.net/lists/listinfo/jython-users > |
From: Dong, M. <MDong@StateStreet.com> - 2006-06-29 15:22:36
|
Hi, I want to write the output of the os.system() to a log file, but don't know how to read the output to a string. Could you give me some suggestion? Many thanks. Ming Dong |
From: Dave K. <dku...@re...> - 2006-06-29 15:57:25
|
On Thu, Jun 29, 2006 at 11:22:21AM -0400, Dong, Ming wrote: > > > Hi, > > I want to write the output of the os.system() to a log file, but > don't know how to read the output to a string. Could you give me some > suggestion? You want one of the variants of popen. See: http://docs.python.org/lib/os-newstreams.html http://docs.python.org/lib/module-popen2.html Here is an example: def test1(): outstream = os.popen('ls *.py', 'r') for line in outstream: print line.rstrip() The other versions (popen2, popen3, ...) enable you to do things like pipe input to the command and read its output, read from stderr as well as stdout, etc. Dave -- Dave Kuhlman http://www.rexx.com/~dkuhlman |
From: Dong, M. <MDong@StateStreet.com> - 2006-06-29 20:20:09
|
When run os.popen("ls","r"), I got error message like: AttributeError: class 'org.python.modules.os' has no attribute 'popen' I am using jython 2.1, any suggestions? Thanks a lot! Ming Dong -----Original Message----- From: jyt...@li... [mailto:jyt...@li...] On Behalf Of Dave Kuhlman Sent: Thursday, June 29, 2006 11:57 AM To: jyt...@li... Subject: Re: [Jython-users] Help: how to get the os.system() stdout On Thu, Jun 29, 2006 at 11:22:21AM -0400, Dong, Ming wrote: >=20 >=20 > Hi, >=20 > I want to write the output of the os.system() to a log file, but > don't know how to read the output to a string. Could you give me some > suggestion? You want one of the variants of popen. See: http://docs.python.org/lib/os-newstreams.html http://docs.python.org/lib/module-popen2.html Here is an example: def test1(): outstream =3D os.popen('ls *.py', 'r') for line in outstream: print line.rstrip() The other versions (popen2, popen3, ...) enable you to do things like pipe input to the command and read its output, read from stderr as well as stdout, etc. Dave --=20 Dave Kuhlman http://www.rexx.com/~dkuhlman Using Tomcat but need to do more? Need to support web services, security? Get stuff done quickly with pre-integrated technology to make your job easier Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo http://sel.as-us.falkag.net/sel?cmd=3Dlnk&kid=3D120709&bid=3D263057&dat=3D= 121642 _______________________________________________ Jython-users mailing list Jyt...@li... https://lists.sourceforge.net/lists/listinfo/jython-users |
From: Raghuram D. <ra...@sy...> - 2006-06-29 20:27:15
|
You can do something like os.system("ls -l >/tmp/testfile"). You may want to generate a temp file name and delete it after reading it. Note that readng stdout/err is very tricky in java as well. Dong, Ming wrote: > When run os.popen("ls","r"), I got error message like: > > AttributeError: class 'org.python.modules.os' has no attribute 'popen' > > I am using jython 2.1, any suggestions? > > Thanks a lot! > > Ming Dong > > -----Original Message----- > From: jyt...@li... > [mailto:jyt...@li...] On Behalf Of Dave > Kuhlman > Sent: Thursday, June 29, 2006 11:57 AM > To: jyt...@li... > Subject: Re: [Jython-users] Help: how to get the os.system() stdout > > On Thu, Jun 29, 2006 at 11:22:21AM -0400, Dong, Ming wrote: >> >> Hi, >> >> I want to write the output of the os.system() to a log file, but >> don't know how to read the output to a string. Could you give me some >> suggestion? > > You want one of the variants of popen. See: > > http://docs.python.org/lib/os-newstreams.html > http://docs.python.org/lib/module-popen2.html > > Here is an example: > > def test1(): > outstream = os.popen('ls *.py', 'r') > for line in outstream: > print line.rstrip() > > The other versions (popen2, popen3, ...) enable you to do things > like pipe input to the command and read its output, read from > stderr as well as stdout, etc. > > Dave > > |
From: Dave K. <dku...@re...> - 2006-06-29 21:36:22
|
On Thu, Jun 29, 2006 at 04:20:02PM -0400, Dong, Ming wrote: > When run os.popen("ls","r"), I got error message like: > > AttributeError: class 'org.python.modules.os' has no attribute 'popen' > > I am using jython 2.1, any suggestions? > OK. Kick me. I apologize. It looks like os.popen is implemented in Jython 2.2a, but not in Jython 2.1. In Jython 2.1, there *is*, however, a popen2 module. But when I tried it, I get the following: $ rlwrap -r ../Jython-2.1/jython Jython 2.1 on java1.4.2_10 (JIT: null) Type "copyright", "credits" or "license" for more information. >>> import popen2 >>> f1, f2 = popen2.popen2('ls -l *.py') Traceback (innermost last): File "<console>", line 1, in ? File "/home/dkuhlman/a1/Python/Jython/Jython-2.1/Lib/popen2.py", line 143, in popen2 File "/home/dkuhlman/a1/Python/Jython/Jython-2.1/Lib/popen2.py", line 35, in __init__ AttributeError: class 'org.python.modules.os' has no attribute 'pipe' But, it works under Jython 2.2a: $ rlwrap -r ../Jython-2.2a/jython Jython 2.2a1 on java1.4.2_10 (JIT: null) Type "copyright", "credits" or "license" for more information. >>> >>> import popen2 >>> f1, f2 = popen2.popen2('ls *.py') >>> f2.close() >>> a = f1.read() >>> print a Foo.py event.py lambda.py o o o Dave -- Dave Kuhlman http://www.rexx.com/~dkuhlman |
From: <cl...@br...> - 2006-06-29 22:01:14
|
Dear colleagues,=20 here the "Java flavor" of OS system access:=20 from java.lang import Runtime from java.io import BufferedReader from java.io import InputStreamReader p =3D Runtime.getRuntime().exec("ping 127.0.0.1")=20 stdInput =3D BufferedReader(InputStreamReader(p.getInputStream())) stdError =3D BufferedReader(InputStreamReader(p.getErrorStream())) print "Here is the standard output of the command:\n" s =3D ' '=20 while (s !=3D None):=20 s =3D stdInput.readLine() print s print "Here is the standard error of the command (if any):\n" s =3D ' '=20 while (s !=3D None): s =3D stdError.readLine()=20 print s=20 Claude Falbriard=20 Developer=20 AMS Hortol=E2ndia / SP - Brazil=20 phone: +55 13 8117 3316 e-mail: cl...@br... "Dong, Ming" <MDong@StateStreet.com>=20 Sent by: jyt...@li... 06/29/2006 12:22 PM To <jyt...@li...> cc Jeff Emanuel <jem...@fr...> Subject [Jython-users] Help: how to get the os.system() stdout Hi, I want to write the output of the os.system() to a log file, but don't know how to read the output to a string. Could you give me some suggestion? Many thanks. Ming Dong Using Tomcat but need to do more? Need to support web services, security? Get stuff done quickly with pre-integrated technology to make your job=20 easier Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo http://sel.as-us.falkag.net/sel?cmd=3Dlnk&kid=3D120709&bid=3D263057&dat=3D1= 21642 =5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F= =5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F Jython-users mailing list Jyt...@li... https://lists.sourceforge.net/lists/listinfo/jython-users |