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.
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