Split file on a pattern

Question

I have a file as below.

***ABC***
Line1*
L*ine2*
***DEF***
Line3*
Lin*e4*

I need to create the 2 files ABC and DEF with the data below:

ABC.dat:

Line1*
L*ine2*

DEF.dat:

Line3*
Lin*e4*

I am ok to generate the file using small shell script as well.

I removed the * you were using since they were confusing the formatting. Are the * part of the file? How can we know which lines are the headers (the file names) and which the content? I mean, what's the difference between ABC and Line1? Is it every 3rd line? Is it lines with ALL CAPS? Please edit your question and clarify. — terdon
– terdon ♦, Commented Apr 6, 2017 at 11:02
* is part of file. Each data line has final selimiter as *. Not every 3rd line. There will be indefinite number of lines between each pattern — Sreekesh Darshi
– Sreekesh Darshi, Commented Apr 6, 2017 at 11:33
OK, next time, please use the formatting tools to format your examples as code. So, can we assume that lines starting with *** will be the target file name? — terdon
– terdon ♦, Commented Apr 6, 2017 at 11:45
Yes correct.... target file name will be the line starting with *** — Sreekesh Darshi
– Sreekesh Darshi, Commented Apr 6, 2017 at 11:51

Sundeep · Accepted Answer · 2017-04-06 12:04:41Z

2

$ awk '/^\*\*\*/{gsub(/\*/,""); fname=$0".dat"; next} {print > fname}' ip.txt 
$ cat ABC.dat 
Line1*
L*ine2*
$ cat DEF.dat 
Line3*
Lin*e4*

/^\*\*\*/ match lines starting with ***
- gsub(/\*/,"") delete all * from such lines
- fname=$0".dat" assign the remaining characters appended with .dat to fname variable
- next move on to next line
print > fname print current line content to filename saved in fname

answered Apr 6, 2017 at 12:04

Sundeep

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