python -i (interactive environment)

[Joe]

When you run "python -i scriptname.py" after the script completes you left
at the interactive command prompt.

Is there a way to have this occur from a running program?

In other words can I just run scriptname.py (NOT python -i scriptname.py)
and inside of scriptname.py I decide that I want to fall back to the
interactive prompt?

I've searched and so far the only thing I've come up with is to use pdb, but
that is not exactly the same as the interactive prompt.

Is there any way to do it that I have missed?

Thanks.

[Joe]

Reinhold,

Interesting.

A key difference between the two is that PYTHONINSPECT will allow you access
to the prompt at the end of your program (assuming no sys.exit or raise
SystemExit) but code.interact() allows you to jump into the program at any
point.
"Reinhold Birkenfeld" <re************ ************@wo lke7.net> wrote in
message news:39******** *****@individua l.net...

/* Check this environment variable at the end, to give programs the
* opportunity to set it from Python.
*/

Reinhold

[Steve Holden]

Michael Hoffman wrote:

Joe wrote:

I want the script to decide whether to fall back to the interactive
prompt. You solution makes it ALWAYS fall back to the interactive prompt.

Actually, using sys.exit() means the program can exit even if python -i
is used.

You can use:

import code
code.interact()

which emulates the interactive prompt.

Unfortunately it does so in an entirely new namespace, thereby losing
the advantage of -i - namely, that you can investigate the program's
namespace after it's terminated.

regards
Steve

[Just]

In article <ma************ *************** *********@pytho n.org>,
Steve Holden <st***@holdenwe b.com> wrote:

Michael Hoffman wrote:

Joe wrote:

I want the script to decide whether to fall back to the interactive
prompt. You solution makes it ALWAYS fall back to the interactive prompt.

Actually, using sys.exit() means the program can exit even if python -i
is used.

You can use:

import code
code.interact()

which emulates the interactive prompt.

Unfortunately it does so in an entirely new namespace, thereby losing
the advantage of -i - namely, that you can investigate the program's
namespace after it's terminated.

code.interact() has a namespace argument ('local'), so it really easy to
have it use the namespace you want.

Just

[Paulo Eduardo Neves]

When I'm using pyunit and want to stop in a point during the test
(skipping all the framework initialization) , I just start the debugger:

import pdb
pdb.set_trace()

You'll get the debugger prompt.

[Joe]

Found that out :-(

You can use the local=locals() option so at least you have access to the
local variables, which in the case of debugging, is exactly what I needed.

Since -i gives you control at the end of the program the locals are already
gone.

Seems like both approaches have their advantages depending on the situation.

"Steve Holden" <st***@holdenwe b.com> wrote in message
news:ma******** *************** *************@p ython.org...

Unfortunately it does so in an entirely new namespace, thereby losing the
advantage of -i - namely, that you can investigate the program's namespace
after it's terminated.

[Joe]

Right, but only one namespace. Would be nice if there was a way to give it
both the global and the local namespaces. In my case though the local
namespace was sufficient.

"Just" <ju**@xs4all.nl > wrote in message
news:ju******** *************** *@news1.news.xs 4all.nl...

code.interact() has a namespace argument ('local'), so it really easy to
have it use the namespace you want.

Just

[Joe]

Isn't this a bug?

Here's the test program:

import code

def test_func():
lv = 1
print '\n\nBEFORE lv: %s\n' % (lv)
code.interact(l ocal=locals())
print '\n\nAFTER lv: %s\n' % (lv)
return

test_func()

gv = 1
print '\n\nBEFORE gv: %s\n' % (gv)
code.interact(l ocal=locals())
print '\n\nAFTER gv: %s\n' % (gv)

Here's the output and interactive session:

BEFORE lv: 1

Python 2.4 (#60, Nov 30 2004, 11:49:19) [MSC v.1310 32 bit (Intel)] on win32
Type "help", "copyright" , "credits" or "license" for more information.
(InteractiveCon sole)

lv = 2
print lv 2 ^Z
AFTER lv: 1

BEFORE gv: 1

Python 2.4 (#60, Nov 30 2004, 11:49:19) [MSC v.1310 32 bit (Intel)] on win32
Type "help", "copyright" , "credits" or "license" for more information.
(InteractiveCon sole) gv = 2
print gv 2 ^Z

AFTER gv: 2

In the case of code.interact being called inside the function, the locals
were available to the interactive environment but changes did not stick when
you returned back the function. (started as lv=1 changed to lv=2 in
interactive session, back to lv=1 in function) Since you are still in the
function and since code.interact used the same local environment why didn't
the change stick until testfunc ended?

When code.interact was called from the outside the function (globals() ==
locals()) the changes stuck. (started as gv=1, changed to gv=2 in
interactive session, stuck as gv=2 back in main).

"Steve Holden" <st***@holdenwe b.com> wrote in message
news:ma******** *************** *************@p ython.org... Michael Hoffman wrote:

Joe wrote:

I want the script to decide whether to fall back to the interactive
prompt. You solution makes it ALWAYS fall back to the interactive
prompt.

Actually, using sys.exit() means the program can exit even if python -i
is used.

You can use:

import code
code.interact()

which emulates the interactive prompt.

Unfortunately it does so in an entirely new namespace, thereby losing the
advantage of -i - namely, that you can investigate the program's namespace
after it's terminated.

regards
Steve

[Steven Bethard]

Joe wrote:

Isn't this a bug?

Here's the test program:

import code

def test_func():
lv = 1
print '\n\nBEFORE lv: %s\n' % (lv)
code.interact(l ocal=locals())
print '\n\nAFTER lv: %s\n' % (lv)
return

Check the documentation for locals() [1]:

"Update and return a dictionary representing the current local symbol
table. Warning: The contents of this dictionary should not be modified;
changes may not affect the values of local variables used by the
interpreter."

So if you change things in the dictionary returned by locals() you won't
actually change the local variables.

STeVe

[1] http://docs.python.org/lib/built-in-funcs.html#l2h-45

[Joe]

Steve,

Thanks, I knew about that but my question is why is it not working
consistently?

Joe
"Steven Bethard" <st************ @gmail.com> wrote in message
news:09******** ************@co mcast.com...

Joe wrote:

Isn't this a bug?

Here's the test program:

import code

def test_func():
lv = 1
print '\n\nBEFORE lv: %s\n' % (lv)
code.interact(l ocal=locals())
print '\n\nAFTER lv: %s\n' % (lv)
return

Check the documentation for locals() [1]:

"Update and return a dictionary representing the current local symbol
table. Warning: The contents of this dictionary should not be modified;
changes may not affect the values of local variables used by the
interpreter."

So if you change things in the dictionary returned by locals() you won't
actually change the local variables.

STeVe

[1] http://docs.python.org/lib/built-in-funcs.html#l2h-45

[Steven Bethard]

Joe wrote:

Thanks, I knew about that but my question is why is it not working
consistently?

At the module level, locals() is globals():

py> locals() is globals()
True

And the globals() dict is modifiable.

HTH,

STeVe