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I wanted to remove white space from a CSV file, which I am able to do using s/\ //g, but at the same time I want to avoid removing the space between fields which has timestamps e.g. "06-JAN-15 13:20:00": currently it joins them which was expected "06-JAN-1513:20:00".

One solution is to let it delete all the white spaces and then grep for date 06-JAN-15 and add space after it. Not so sure how that can be done.

Sample CSV file: (one line only)

294335,"17-APR-15 00:00:00 ",6258,"C"," <-----huge blank space--->,"07-JAN-15 00:00:00"

huge blank space will contain XML message if not blank.

  • Crossposting: stackoverflow.com/questions/31145618/… – Cyrus Jun 30 '15 at 19:06
  • Please provide some example lines and your desired output from them.. – heemayl Jun 30 '15 at 19:18
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    are you just trying to remove large sections of contiguous whitespace? Maybe 's/ */ /g' (space-space-star) or 's/ \+/ /g' to replace 2+ spaces with 1 would be "good enough"? – drewbenn Jun 30 '15 at 19:43
  • ... or are you just trying to collapse any entirely blank fields? – steeldriver Jun 30 '15 at 19:54
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    @drewbenn adding " s/ \+/ /g " removed the blank space with single space. Thanks a lot new output 294335,"11-JUN-15 00:00:00 ",0,"N"," ",30-JUN-15 13:30:00 " – JavaQuest Jun 30 '15 at 20:15
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Not sed, but here's a solution using negative lookbehind assertion (?<!...) in perlre.

perl -pe 's/(?<!\d\d-(JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV|DEC)-\d\d) //g'
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Using awk:

< input awk 'BEGIN {FS=OFS=","} {gsub(" ", "", $1); gsub(" ", "", $3); gsub(" ", "", $4); gsub(" ", "", $5); print}'

This will have the effect of removing any space character from the 1st, 3rd, 4th and 5th comma-separated field in each record.

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You could actually parse the file as a CSV file, and strip extra whitespace:

ruby -rcsv -ne '
    row = CSV::parse_line($_)
    puts CSV::generate_line(row.map {|field| field.strip})
' file
294335,17-APR-15 00:00:00,6258,C,"",07-JAN-15 00:00:00

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