Bash parameter substitution question

kushalkoolwal · 03-22-2011, 05:29 PM

Let's say I have a following string in a variable called LONG_NAME:

Code:

ata-WDC_WD3200AAJB-00J3A0_WD-WCAV2Y020369 wwn-0x50014ee103971beb

Now I want to simply take out everything in the red color above.

I have been trying the following command but it does not work:

Code:

echo "${LONG_NAME%%_[0-9]*}"

Basically, I am trying to remove anything after the second underscore (including).

jlinkels · 03-22-2011, 06:25 PM

Code:

jlinkels@donald_pc:~$ echo ${LONG_STRING%_[A-Z0-9]*}
ata-WDC_WD3200AAJB-00J3A0

Provided there are no more underscores in the part that you want to remove. The trick is to match the shortest string (not the longest) so you can be more specific.

jlinkels

ntubski · 03-22-2011, 06:31 PM

It looks like maybe you thought ${var%%pat} expansion uses regex, but actually the pattern is a glob, see Shell Parameter Expansion

Code:

${LONG_NAME%_*}

would remove everything after the last underscore.

kurumi · 03-22-2011, 07:33 PM

Code:

$ s="ata-WDC_WD3200AAJB-00J3A0_WD-WCAV2Y020369 wwn-0x50014ee103971beb"
$ echo $s| ruby -e 'puts gets.split(/_/)[0,2].join("_")'
ata-WDC_WD3200AAJB-00J3A0

kushalkoolwal · 03-29-2011, 12:16 PM

Thank you all. All the suggestions were good.

grail · 03-29-2011, 08:36 PM

Don't forget to mark as SOLVED once you have a solution