I have a text file contaning the following:


The separator is "

How can I split the containt into multiple file as follow using a bash command?

File 1 :


File 2 :


I came into examples using csplit or awk but they does not cover this text layout.

  • Is that the only text in the file? – Hauke Laging Dec 8 '14 at 17:15
  • yes only text in the file – Fabio Dec 9 '14 at 9:38

Simple awk command:

awk 'NR%2==0{ print > "File "++i }' RS='"' file

RS defines " as record separator and NR is the record number. If the record number was modulo of 2 (because we have another first " for records), then print the current record $0 into a File #.

  • 1
    This is IMO the best answer as it works even if other lines exist between "$ and the next ^". – don_crissti Dec 8 '14 at 21:26
  • don't need the begin just use ++i instead of i++.Could be shortened to awk 'NR%2==0{ print >"File "++i }' RS='"' file – user78605 Dec 8 '14 at 22:33
  • @Jidder Uh,yes. thank you. updated with that, – αғsнιη Dec 9 '14 at 4:40

If the opening quote is always at the beginning of the line, csplit will work just fine like this:

bash$ csplit /tmp/data '/^"/'

That produces files called xx00, etc. Note, your example removes the quotation marks and this doesn't. You'd need a command line sed to do that:

bash$ for file in xx* ; do { sed 's/^"//;s/"$//;' ${file} >x${file}; } ; done
  • With coreutils csplit: csplit -z filesample '/^"/' {*} so as to split more than once and ignore the first piece (xx00) which will always be empty (for the given text sample). – don_crissti Dec 8 '14 at 18:20

With GNU awk

awk -v RS='"[[:space:]]*"' '
{sub(/^"|"[[:space:]]*$/, "");print > "output." ++n; close("output." n)}' file.txt
  • An alternative: awk -v RS='"' '!/^[[:space:]]*$/ {print > "output." n++; close...}' – muru Dec 8 '14 at 17:49

You can do:

awk -v RS=\" -v ORS= \
  '{ sub(/^\n*/,"");sub(/\n*$/,""); };
  /^\n*$/ { next;};
  { if(strcnt==0) { print $0 >"file1"; strcnt++;} else
  { print $0 >"file2"; }; }' file

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