parsing pyton code

[Philippe C. Martin]

Hi,

I'm looking for a very simple way to figure out if a line of code requires
further lines of code ex: for i in range(10): So far I have found the
following trick that seem dangerous to me since it assumes the string 'EOF'
exists in the exception message.

Do I have the right to do this ?




import compiler
import string
try:
compiler.parse( 'for i in range(10):')
except SyntaxError, (val1, val2):
print val1
print val2
l_eof = False
l = string.split(va l1,' ')
for l_c in l:
if l_c == 'EOF':
l_eof = True
break
if l_eof == True:
print 'YES'

[Jean Brouwers]

Re: parsing pyton code


Check the source of the 'code' module, especially the runsource()
method of the InteractiveInte rpreter class. Also, see these docs

<http://docs.python.org/lib/module-code.html>


/Jean Brouwers
ProphICy Semiconductor, Inc.


<mailman.6015.1 099773489.5135. python-list@python.org >, Philippe C.
Martin <philippecmarti n@sbcglobal.net > wrote:
[color=blue]
> Hi,
>
> I'm looking for a very simple way to figure out if a line of code requires
> further lines of code ex: for i in range(10): So far I have found the
> following trick that seem dangerous to me since it assumes the string 'EOF'
> exists in the exception message.
>
> Do I have the right to do this ?
>
>
>
>
> import compiler
> import string
> try:
> compiler.parse( 'for i in range(10):')
> except SyntaxError, (val1, val2):
> print val1
> print val2
> l_eof = False
> l = string.split(va l1,' ')
> for l_c in l:
> if l_c == 'EOF':
> l_eof = True
> break
> if l_eof == True:
> print 'YES'[/color]