How to extract a part of file name in unix/linux shell script

Question

My file name is Oracle.2347263_testing_152607.csv I tried this :

echo Oracle.2347263_testing_152607.csv | sed -e 's/.*G.//'   \-e 's/.csv//'

but I didn't get the result I expected.

I want to extract 2347263 from Oracle.2347263_testing_152607.csv please help

If you're using bash and the filename is already in a variable, this is cheaper: i=Oracle.2347263_testing_152607.csv and then j=${i#Oracle.*} ; echo ${j%%_*}. — ott--, Feb 3 '15 at 23:06

John1024 · Accepted Answer · 2015-02-02 23:55:42Z

1

$ echo Oracle.2347263_testing_152607.csv | sed -e 's/Oracle.\([[:digit:]]*\)_.*/\1/'
2347263

answered Feb 2 '15 at 23:44

John1024

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Thanks for the response. I tried this but got some error sed: illegal option -- r Usage: sed [-n] [-u] Script [File ...] sed [-n] [-u] [-e Script] ... [-f Script_file] ... [File ...] – ora_unix Feb 2 '15 at 23:46
"got some error" What error? It works fine for me. If it doesn't work for you, show me exactly what happened. – John1024 Feb 2 '15 at 23:46
sed: illegal option -- r Usage: sed [-n] [-u] Script [File ...] sed [-n] [-u] [-e Script] ... [-f Script_file] ... [File ...] – ora_unix Feb 2 '15 at 23:48
@ora_unix OK. That error looks like you are using a BSD sed. Answer updated. – John1024 Feb 2 '15 at 23:56

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slm · Accepted Answer · 2015-02-03 22:02:48Z

0

Try the following:

$ echo Oracle.2347263_testing_152607.csv |cut -f 2 -d "." |cut -f 1 -d "_"
2347263

slm♦

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answered Feb 3 '15 at 21:31

undefine

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2 Answers 2