Using bash: how to split file according to input file line content?

Question

I've got some big text log files. The content is like:

Begin to work
Load library
Start
TEXTLOG
Checking
ok
TEXTLOG
Start process
Starting node
ok
TEXTLOG
Stop node
TEXTLOG

In this file the lines "TEXTLOG" serves as delimeter, so I wish to split this file into several smaller files, using "TEXTLOG" as EOF indicator, so I should files:

file1:

Begin to work
Load library
Start

file2:

Checking
ok

file3:

Start process
Starting node
ok

file4:

Stop node

How can I achieve this using shell? Thanks.

oliv · Accepted Answer · 2018-08-23 11:40:30Z

2

Using GNU awk:

awk -v RS="\nTEXTLOG\n" '{print > "file"++c}' file

RS is the record separator and allows to get the content of each file at once.

answered Aug 23, 2018 at 11:40

oliv

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Mats Faugli · Accepted Answer · 2018-08-23 11:57:23Z

2

Using csplit:

csplit --suppress-matched file "/TEXTLOG/" "{*}"

--suppress-matched makes sure the TEXTLOG separator is not included in the output
"{*}" tells csplit to run the matching pattern not just once, but throughout the entire file

answered Aug 23, 2018 at 11:57

Mats Faugli

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2 Comments

Hind Forsum Over a year ago

seems my csplit on linux/mac doesn't support --suppress-matches. How did you try?

Mats Faugli Over a year ago

I'm running Ubuntu 18.04. csplit --version gives me csplit (GNU coreutils) 8.28.

Gilles Quénot · Accepted Answer · 2018-08-23 10:37:49Z

1

Try this :

awk 'BEGIN{c=1}/^TEXTLOG/{c++;next} {print > "file"c}' file 
ls

answered Aug 23, 2018 at 10:37

Gilles Quénot

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