Shell program, error in for loop. Variable not used

Question

I am quite new to shell scripting. My code should be very simple, quite a tutorial, but I cannot make it work:

case='Naca0012_pitch_V1'
for i in {0..800..10} 
do
    tec $case.$i.plt;
    cat tec.00* > $case.$i.dat;
done

The command tec generates some files that I need to concatenate.

What I obtain is:

Error opening file Naca0012_pitch_V1.{0..800..10}.plt

from the tec utility.

It seems that the code puts "{0..800..10}" instead of substituting the variable with one of the possible cases from the brace expansion.

What is the shebang of your script, which shell are you using? You mention bash, but this is perfect valid bash. — kvantour
– kvantour, Commented Dec 13, 2018 at 9:23
It's perfect bash if you are on version 4.0+ Please verify your bash version with echo $BASH_VERSION — MadJlzz
– MadJlzz, Commented Dec 13, 2018 at 9:31
In terminal, do ls -l /bin/sh. Maybe your shell is /bin/dash ?? — user2849202
– user2849202, Commented Dec 13, 2018 at 9:31

kvantour · Accepted Answer · 2018-12-13 09:45:33Z

2

You try to make use of brace expansion. Depending on the shell you are using, brace expansion is supported or not. Shells that do not support brace expansion are POSIX sh and dash. Shells that do support braces expansion are bash, ksh and zsh.

Posix complient shells (sh or dash): If you want to mimick this kind of brace expansion in sh or dash, you have to do something like

for i in `seq 0 10 800`; do
  ...
done

Disable brace expansion: While bash, ksh and zsh all support brace expansion, it is possible that it has been disabled. For bash and ksh this can be done by executing the command

$ set +B
$ for i in {0..800..10}; do echo $i; done
{0..800..10}

edited Dec 13, 2018 at 9:45

answered Dec 13, 2018 at 9:30

kvantour

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1 Comment

federico rovere Over a year ago

Thank you for explanation. I was using the wrong shell.

Derviş Kayımbaşıoğlu · Accepted Answer · 2018-12-13 09:21:19Z

-1

Your loop ia not correct.

If your intention is looping from 0 to let say 800. Your loop is needed to be like

for n in $(seq 1 100)
do 
    #yourcode 
done

answered Dec 13, 2018 at 9:21

Derviş Kayımbaşıoğlu

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1 Comment

kvantour Over a year ago

Sorry but your statement is wrong, the loop is perfectly valid.

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