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Re: Help on PyQt4 QProcess

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On Fri, 19 Aug 2011 14:32:12 -0700 (PDT), Edgar Fuentes
<fuentesej@gmail.com> wrote:
> On Aug 19, 4:21 pm, Carl Banks <pavlovevide...@gmail.com> wrote:
>> On Friday, August 19, 2011 12:55:40 PM UTC-7, Edgar Fuentes wrote:
>> > On Aug 19, 1:56 pm, Phil Thompson
>> >  wrote:
>> > > On Fri, 19 Aug 2011 10:15:20 -0700 (PDT), Edgar Fuentes
>> > > <fuen...@gmail.com> wrote:
>> > > > Dear friends,
>>
>> > > > I need execute an external program from a gui using PyQt4, to
avoid
>> > > > that hang the main thread, i must connect the signal
>> > > > "finished(int)"
>> > > > of a QProcess to work properly.
>>
>> > > > for example, why this program don't work?
>>
>> > > >    from PyQt4.QtCore import QProcess
>> > > >    pro = QProcess() # create QProcess object
>> > > >    pro.connect(pro, SIGNAL('started()'), lambda
>> > > > x="started":print(x))        # connect
>> > > >    pro.connect(pro, SIGNAL("finished(int)"), lambda
>> > > > x="finished":print(x))
>> > > >    pro.start('python',['hello.py'])        # star hello.py
program
>> > > > (contain print("hello world!"))
>> > > >    timeout = -1
>> > > >    pro.waitForFinished(timeout)
>> > > >    print(pro.readAllStandardOutput().data())
>>
>> > > > output:
>>
>> > > >    started
>> > > >    0
>> > > >    b'hello world!\n'
>>
>> > > > see that not emit the signal finished(int)
>>
>> > > Yes it is, and your lambda slot is printing "0" which is the return
>> > > code
>> > > of the process.
>>
>> > > Phil
>>
>> > Ok, but the output should be:
>>
>> >     started
>> >     b'hello world!\n'
>> >     finished
>>
>> > no?.
>>
>> > thanks Phil
>>
>> Two issues.  First of all, your slot for the finished function does not
>> have the correct prototype, and it's accidentally not throwing an
>> exception because of your unnecessary use of default arguments.
 Anyway,
>> to fix that, try this:
>>
>> pro.connect(pro, SIGNAL("finished(int)"), lambda v,
>> x="finished":print(x))
>>
>> Notice that it adds an argument to the lambda (v) that accepts the int
>> argument of the signal.  If you don't have that argument there, the int
>> argument goes into x, which is why Python prints 0 instead of
"finished".
>>
>> Second, processess run asynchrously, and because of line-buffering, IO
>> can output asynchronously, and so there's no guarantee what order
output
>> occurs.  You might try calling the python subprocess with the '-u'
switch
>> to force unbuffered IO, which might be enough to force synchronous
output
>> (depending on how signal/slot and subprocess semantics are
implemented).
>>
>> Carl Banks
> 
> Thanks Carl, your intervention was very helpful for me, this solve my
> semantic error. I need to study more about signal/slots and process.

In which case you should look at the modern, Pythonic connection syntax
rather than the old one...

    pro.started.connect(lambda: print("started"))
    pro.finished.connect(lambda: print("finished"))

Phil

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Thread

Help on PyQt4 QProcess Edgar Fuentes <fuentesej@gmail.com> - 2011-08-19 10:15 -0700
  Re: Help on PyQt4 QProcess Phil Thompson <phil@riverbankcomputing.com> - 2011-08-19 18:56 +0100
    Re: Help on PyQt4 QProcess Edgar Fuentes <fuentesej@gmail.com> - 2011-08-19 12:55 -0700
      Re: Help on PyQt4 QProcess Carl Banks <pavlovevidence@gmail.com> - 2011-08-19 13:21 -0700
        Re: Help on PyQt4 QProcess Edgar Fuentes <fuentesej@gmail.com> - 2011-08-19 14:32 -0700
          Re: Help on PyQt4 QProcess Phil Thompson <phil@riverbankcomputing.com> - 2011-08-20 09:36 +0100
            Re: Help on PyQt4 QProcess Edgar Fuentes <fuentesej@gmail.com> - 2011-08-22 10:43 -0700

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