execfile() on file subclass or string

[Brano Zarnovican]

Hi !

I have a python script represented by a string.
I need to execute it in a context. 'exec' does
the job, but doesn't display the filename in
tracebacks. 'execfile' is displaying the filename
but it can only exec a script in a filesystem.

I have tried:
- to give exec a filename, like:

exec script_content in dict({'__file__ ':'bla.py'})

=> didn't work (neighter with __name__)

- create an in-memory file subclassing 'file'

class memfile(file):
..
def read(self)..

=> also pointless as 'execfile' doesn't accept a file
only a filename

Can you help me ?
The only problem I need to resolve is that I don't see
the script name in tracebacks.

Thanks,

BranoZ

PS: The script didn't came from the filesystem, but I know
it's content and a name.

[Do Re Mi chel La Si Do]

Hi !
little idea :
You can save the string, in temp-directory (see tempfile module), then,
use execfile
Michel Claveau

[Fredrik Lundh]

Brano Zarnovican wrote:

I have a python script represented by a string.
I need to execute it in a context. 'exec' does
the job, but doesn't display the filename in
tracebacks. 'execfile' is displaying the filename
but it can only exec a script in a filesystem.

I have tried:
- to give exec a filename, like:

exec script_content in dict({'__file__ ':'bla.py'})

=> didn't work (neighter with __name__)

compile the script first, and execute the resulting code object:

exec compile("code", "filename", "exec")

Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "filename", line 1, in ?
NameError: name 'code' is not defined

</F>

[Brano Zarnovican]

> exec compile("code", "filename", "exec")

Thanks for the tip!
Works great!

BranoZ