Character replacement


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# 1  
Character replacement

Hi,

I am working on a command that replaces some occurrences of quotation marks in file. The quotation mark cannot be the first or the last character in line and cannot be preceded or followed by a comma.

I am not an expert in regular expressions, but I managed to create the following command.

sed 's/\([^,]\)\"\([^,]\)/\1\"\"\2/g' file.txt

Can you please verify it or tell me how to do it better? Any help will be appreciated.

Regards,
Pit
# 2  
Hi,

Small update regarding my problem. Currently I need to match pattern from the last occurence of
Code:
","

or
Code:
,,"

till the end of line. For example matching on the following line:

Code:
"one","two","three",,"five",,"sev"e,n"

should return
Code:
sev"e,n

. I would like to use sed or awk, but any other solution will be fine. Can you please help me writing a proper regular expression?

Pit
# 3  
you could do within vi the global repleace

:g/c2/s//c5/g

this replaces everything with c2 to c5. you could some how work this into a script to do a larger directory if you do not feel like vi a file. not sure though
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