FrEaK85
asked on
Shell programing and regular expressiong: -> grep
Ok, what I need to do is: list all files in a given directory with a given extension
$1 = directory
$2 = extension
I want to use regular expression for this, so I used the command
filelist = `ls $1 | grep '.*[\\\\.]$2$'`
the problem is, the $ from $2 conflicts with the $ from the regular expression
What I've already tried:
- putting the command in a dummy variable as a string and then execute it
- putting all sort of quotes around it
- banging my head on the table
PS: it works if I replace $2 with the actual extension
This is a school assignment so I can't use any other language or so
Hope someone can help me,
thanks,
Walter
$1 = directory
$2 = extension
I want to use regular expression for this, so I used the command
filelist = `ls $1 | grep '.*[\\\\.]$2$'`
the problem is, the $ from $2 conflicts with the $ from the regular expression
What I've already tried:
- putting the command in a dummy variable as a string and then execute it
- putting all sort of quotes around it
- banging my head on the table
PS: it works if I replace $2 with the actual extension
This is a school assignment so I can't use any other language or so
Hope someone can help me,
thanks,
Walter
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shivsa
filelist = `ls $1 | grep ".*[\\\\.]$2$''`
also after setting $1 and $2, do not forget to export them.
and run the above command, it is working onmy system.
and run the above command, it is working onmy system.
did ot work, please update.
Patience.
Still don't see why you need to invoke grep when ls was written to do what you want.
Still don't see why you need to invoke grep when ls was written to do what you want.
try it using eval before grep as follows
var1=directory
var2=extension
filelist=`ls | eval grep '.*[\\\\.]$var2$'`
I think, this should work
rgds,
Sanjeev
var1=directory
var2=extension
filelist=`ls | eval grep '.*[\\\\.]$var2$'`
I think, this should work
rgds,
Sanjeev
You don't need to use grep, just simply use:
filelist=`/bin/ls ${1}/*.${2}`
PS: use the full path to ls (do a which ls to find out), in case you have
set aliase for ls.
filelist=`/bin/ls ${1}/*.${2}`
PS: use the full path to ls (do a which ls to find out), in case you have
set aliase for ls.
Actually, there's no need to use 'ls', either. Expanding wildcards (sometimes called globbing) is something the shell does and in all of these formulations using ls, it's really the shell doing the work.
filelist=$(echo $1/*.$2)
As for the way to get your original formulation to work, you need to protect the $ that was intended to be an end-of-string marker from being prematurely taken by the shell as a variable substitution. You seemed to have some concept of this given the number of backslashes you put before that . in the square brackets...but here's a version that worked for me:
filelist=`ls $1 | egrep ".*\\.$2\$"`
The various shells differ a bit in whether variable interpolation takes place inside single quotes inside backticks. To be safe, I stuck to double-quotes.
filelist=$(echo $1/*.$2)
As for the way to get your original formulation to work, you need to protect the $ that was intended to be an end-of-string marker from being prematurely taken by the shell as a variable substitution. You seemed to have some concept of this given the number of backslashes you put before that . in the square brackets...but here's a version that worked for me:
filelist=`ls $1 | egrep ".*\\.$2\$"`
The various shells differ a bit in whether variable interpolation takes place inside single quotes inside backticks. To be safe, I stuck to double-quotes.
ASKER
filelist=`ls $1/*.$2` => feeling stupid stupid stupid