how to improve simple python shell script (to compile list of files)

[Jari Aalto]

[Keep CC, thank you]

Please suggest comments how can I make this script to work
from bash. Also how can I skip better the [0] argument from
command line without hte extra variable i?

#!/bin/bash

function compile ()
{
python -c '
import os, sys, py_compile;
i = 0;
for arg in sys.argv:
file = os.path.basename(arg);
dir = os.path.dirname(arg);
i += 1;
if i > 1 and os.path.exists(dir):
os.chdir(dir);
print "compiling %s\n" % (file);
py_compile.compile(file);
' $*
}

compile $(find path/to -type f -name "*.py")

# End of example

The error message reads:

File "<string>", line 2
import os, sys, py_compile;
^
SyntaxError: invalid syntax

Jari

[Maarten van Reeuwijk]

Jari Aalto wrote:

[Keep CC, thank you]

Please suggest comments how can I make this script to work
from bash. Also how can I skip better the [0] argument from
command line without hte extra variable i?

Didn't check, but something like this?

#!/bin/python
import os, sys, py_compile;
i = 0;
for arg in sys.argv:
file = os.path.basename(arg);
dir = os.path.dirname(arg);
i += 1;
if i > 1 and os.path.exists(dir):
os.chdir(dir);
print "compiling %s\n" % (file);
py_compile.compile(file);

--
================================================== =================
Maarten van Reeuwijk dept. of Multiscale Physics
Phd student Faculty of Applied Sciences
maarten.ws.tn.tudelft.nl Delft University of Technology

[Chris F.A. Johnson]

On 2005-10-15, Jari Aalto wrote:

[Keep CC, thank you]

Please suggest comments how can I make this script to work
from bash. Also how can I skip better the [0] argument from
command line without hte extra variable i?

#!/bin/bash

function compile ()
{
python -c '
import os, sys, py_compile;
i = 0;
for arg in sys.argv:
file = os.path.basename(arg);
dir = os.path.dirname(arg);
i += 1;
if i > 1 and os.path.exists(dir):
os.chdir(dir);
print "compiling %s\n" % (file);
py_compile.compile(file);
' $*
}
Don't indent:

function compile ()
{
python -c '
import os, sys, py_compile;
i = 0;
for arg in sys.argv:
file = os.path.basename(arg);
dir = os.path.dirname(arg);
i += 1;
if i > 1 and os.path.exists(dir):
os.chdir(dir);
print "compiling %s\n" % (file);
py_compile.compile(file);
' $*
}
compile $(find path/to -type f -name "*.py")

# End of example

The error message reads:

File "<string>", line 2
import os, sys, py_compile;
^
SyntaxError: invalid syntax

Jari

--
Chris F.A. Johnson <http://cfaj.freeshell.org>
================================================== ================
Shell Scripting Recipes: A Problem-Solution Approach, 2005, Apress
<http://www.torfree.net/~chris/books/cfaj/ssr.html>

[Jari Aalto]

"Chris F.A. Johnson" <cf********@gmail.com> writes:

On 2005-10-15, Jari Aalto wrote:
Don't indent:

function compile ()
{
python -c '
import os, sys, py_compile;
i = 0;
for arg in sys.argv:
file = os.path.basename(arg);
dir = os.path.dirname(arg);
i += 1;
if i > 1 and os.path.exists(dir):
os.chdir(dir);
print "compiling %s\n" % (file);
py_compile.compile(file);
' $*
}

Thanks, is there equivalent to this Perl statement in Python?

@list = @ARGV[1 .. @ARGV];

or something similar so that I could avoid the 1 > 1 (sys.argv) check
altogether?

Jari

[Robert Kern]

Jari Aalto wrote:

Thanks, is there equivalent to this Perl statement in Python?

@list = @ARGV[1 .. @ARGV];

or something similar so that I could avoid the 1 > 1 (sys.argv) check
altogether?

for arg in sys.argv[1:]:
...

--
Robert Kern
rk***@ucsd.edu

"In the fields of hell where the grass grows high
Are the graves of dreams allowed to die."
-- Richard Harter