string matching in bash 2.0.5

texaganian · 11-30-2009, 09:25 AM

OK, sorry, I know this is pretty basic but my brain doesn't want to wrap around it.

I need to create a bash shell script that tests the first parameter to see if it matches the following regex:

^[rRsS][0-9]{3}

Problem is, the box runs bash 2.0.5 and it doesn't have the regex match operator.

So what's the cleanest way to do this?

pixellany · 11-30-2009, 09:39 AM

1. "first parameter" of what?

2. What is the "regex match operator"?

3. Why not install more current BASH?

You can use that Regex with any number of utilities, including SED and GREP.

texaganian · 11-30-2009, 09:54 AM

Quote:

Originally Posted by pixellany

1. "first parameter" of what?

2. What is the "regex match operator"?

3. Why not install more current BASH?

You can use that Regex with any number of utilities, including SED and GREP.

I need to test the first argument passed to the script. i.e, $1.
The regex match operator I was referring to is the =~ operator in bash 3. That would make it as easy as
Code:
```
if [[ "$1" =~ ^[rRsS][0-9]{3} ]]
```
but...
I can't do that because I don't manage the box so I can't change the installed version of bash.

catkin · 11-30-2009, 10:01 AM

Bash Conditional Expressions do not include a "regex match operator".

Assuming the regular expression you gave means you want to match a string that begins with a character from rRsS followed by 3 characters from 0-9 ... here's an untested way to test $1 for it

Code:

case "$1" in
    [rRsS][0-9][0-9][0-9]* )
        echo 'Got it!'
esac

EDIT: but the [[ ]] detailed in Conditional Constructs does include =~

texaganian · 11-30-2009, 10:34 AM

Quote:

Originally Posted by catkin

Code:

case "$1" in
    [rRsS][0-9][0-9][0-9]* )
        echo 'Got it!'
esac

Perfect! I knew it was going to be pretty straightforward but my brain just wasn't cooperating!