09-30-2010, 06:39 AM
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#4
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Junior Member
Posts: 3
Karma: 10
Join Date: Sep 2010
Location: Brisbane, AU
Device: Kindle
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Quote:
Originally Posted by TonytheBookworm
maybe this will work:
Code:
return url.replace('\"', '\%22)
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Tried it, and got this:
Quote:
Python function terminated unexpectedly: ("'return' outside function", ('/var/folders/ry/ry9uM5AmH88g7YprtmDiOE+++TI/-Tmp-/calibre_0.7.20_tmp_id3zFW/calibre_0.7.20_O6WgL4_recipes/recipe0.py', 27, None, 'return url.replace(\'\\"\', \'\\%22\')\n'))
Traceback (most recent call last):
File "/Applications/calibre.app/Contents/Resources/Python/lib/python2.6/site.py", line 147, in main
return run_entry_point()
File "/Applications/calibre.app/Contents/Resources/Python/lib/python2.6/site.py", line 116, in run_entry_point
return getattr(pmod, func)()
File "site-packages/calibre/ebooks/conversion/cli.py", line 254, in main
File "site-packages/calibre/ebooks/conversion/plumber.py", line 832, in run
File "site-packages/calibre/customize/conversion.py", line 211, in __call__
File "site-packages/calibre/web/feeds/input.py", line 68, in convert
File "site-packages/calibre/web/feeds/recipes/__init__.py", line 47, in compile_recipe
File "/var/folders/ry/ry9uM5AmH88g7YprtmDiOE+++TI/-Tmp-/calibre_0.7.20_tmp_id3zFW/calibre_0.7.20_O6WgL4_recipes/recipe0.py", line 27
return url.replace('\"', '\%22')
SyntaxError: 'return' outside function
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