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Top Forums Shell Programming and Scripting regular expression grepping lines with VARIOUS number of blanks Post 302342647 by ABE2202 on Monday 10th of August 2009 11:02:06 AM
Old 08-10-2009
regular expression grepping lines with VARIOUS number of blanks

Hi,

I need a regular expression grepping all lines starting with '*' followed by a VARIOUS number of blanks and then followed by the string 'Runjob=1'.

I tried that code, but it doesn't work:

Code:
 
grep -i '*'[ ]+'Runjob=1' INPUT_FILE >>OUTPUT_FILE

Can someone help me?

Thanks
 

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MARKDOWN_PY(1)						      General Commands Manual						    MARKDOWN_PY(1)

NAME
markdown_py - a Python implementation of John Gruber's Markdown. SYNOPSIS
markdown_py [options] [INPUT_FILE] (STDIN is assumed if no INPUT_FILE is given) OPTIONS
--version Show program's version number and exit. -h, --help Show the help message and exit. -f OUTPUT_FILE, --file=OUTPUT_FILE Write output to OUTPUT_FILE. Defaults to STDOUT. -e ENCODING, --encoding=ENCODING Encoding for input and output files. -q, --quiet Suppress all warnings. -v, --verbose Print all warnings. -s SAFE_MODE, --safe=SAFE_MODE 'replace', 'remove' or 'escape' HTML tags in input. -o OUTPUT_FORMAT, --output_format=OUTPUT_FORMAT 'xhtml1' (default), 'html4' or 'html5'. --noisy Print debug messages. -x EXTENSION, --extension=EXTENSION Load extension EXTENSION. -n, --no_lazy_ol Observe number of first item of ordered lists. DESCRIPTION
Command-line Markdown compiler based on python-markdown module, with extensions support. AUTHOR
This manual page was written by Dmitry Shachnev <mitya57@gmail.com>. MARKDOWN_PY(1)
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