String substitute in Shell script

Question

I suppose to strip down a substring in my shell script. I am trying as follows:

fileName="Test_VSS_TT.csv.old"

here i want to remove the string ".csv.old" and my

test=${fileName%.*}

but getting bad substitution error.

Did you try `test=${filename%%.*}`? `%%` returns the longest matching pattern. — Fazlin, Sep 30 '14 at 06:30
But the command you tried should have removed the `.old` from your string and resulted in `TEST_VSS_TT.csv` instead of throwing the error, Have you double checked your command ? , — Ram, Sep 30 '14 at 06:40
My first problem is still getting same error even with %%. My system is Solaris sparc. — CrazyC, Sep 30 '14 at 06:44

Avinash Raj · Answer 1 · 2014-09-30T06:40:35.110

0

Here you go,

$ echo $f
Test_VSS_TT.csv.old
$ test=${f%%.*}
$ echo $test
Test_VSS_TT

%% will do a longest match. So it matches from the first dot upto the last and then removes the matched characters.

edited Sep 30 '14 at 06:40

answered Sep 30 '14 at 06:35

Avinash Raj

My first problem is still getting same error even with %%. My system is Solaris sparc – CrazyC Sep 30 '14 at 06:46

Alexandre Halm · Accepted Answer · 2014-09-30T09:15:11.563

0

you are looking for test=${filename%%.*}

the doc for parameter expansion in bash here and in zsh here

%.* will match the first .* pattern, whereas %%.* will match the longest one

[edit]

if sed is available, you could try something like that : echo "filename.txt.bin" | sed "s/\..*//g" which yields filename

edited Sep 30 '14 at 09:15

answered Sep 30 '14 at 06:36

Alexandre Halm

@CrazyC: what shell/version are you using ? – Alexandre Halm Sep 30 '14 at 08:25
Bash version is 3.2.51(1)-release – CrazyC Sep 30 '14 at 08:59
strange. Can you please post a complete copy of the command and the error message from an interactive shell? – Alexandre Halm Sep 30 '14 at 09:02
./mytest.sh : bad substitution. If i comment out that line then no error. – CrazyC Sep 30 '14 at 09:52
1

are you sure your sh is bash ? there seems to be a precedent on StackOverflow [here](http://stackoverflow.com/questions/20615217/bash-bad-substitution). Also, have you tried launching your script with `bash ./mytest.sh` ? – Alexandre Halm Sep 30 '14 at 10:41

score 0 · Answer 3 · answered Sep 30 '14 at 08:33

0

If your intention is to extract file name without extension, then how about this?

$ echo ${fileName}               
Test_VSS_TT.csv.old
$ test=`echo ${fileName} |cut -d '.' -f1`
$ echo $test
Test_VSS_TT

answered Sep 30 '14 at 08:33

akira

score 0 · Answer 4 · answered Sep 30 '14 at 09:50

0

echo "Test_VSS_TT.csv.old"| awk -F"." '{print $1}'

answered Sep 30 '14 at 09:50

dev

simplest way , save the result in a variable and use it when needed – dev Sep 30 '14 at 09:51
awk:syntax error near line 1 awk: bailing out near line 1 – CrazyC Sep 30 '14 at 10:14

4 Answers4