Shell Programming and Scripting

Hi Gurus,


I have a requirement which needs to pass a parameter when calling the script, using this parameter to find a file name stored in master file. then read the file content. the issue is in the file name has a special character "$". don't know how to handle this. below is example:
the master file is:

Code:

abc, v$xyz
bcd, x$efg

my script like:

Code:

#!/bin/ksh


genfile=$1


while IFS=','  filedir, name
do
cd $filedir
if [ -f ${genfile}* ]; then
cat $name ... "read the file and do something. 
fi
done < Masterfile

the real file name like v$xyz_xxxx_xxxx.xxx


I call the script: ksh scriptname 'v$xyz'.


without "$", the script. with "$", I can put single quote when call the script, but in the script, I don't know how to add single quote in variable "name". please share you idea with me how to handle this.


I know I need to wrap the code with code tag, for some reason, the code tag doesn't work for me.


thanks in advance.

What is your browser? If the button doesn't work, you can always do it by yourself, [code]stuff[/code]

What is the script supposed to do with $xyz, etc? Anything? variable xyz doesn't appear in your script.

Your loop is slightly wrong:

Code:

while IFS=", " read filedir name
do
        echo "filedir [$filedir] name [$name]"
done < masterfile

thanks Corona688.
the variable is "v$xyz", the file name like v$xyz_xxxx_xxxx.txt.
basically, I want to use variable "v$xyz" to find "v$xyz_xxxx_xxxx.txt" in the directory, if the file exist, then copy file to ABCD_v$xyz.txt. the directory name saved in master file with .

Code:

 abc, v$xyz
bcd, x$efg

'name' is the variable, 'v$xyz' is the value it holds.

The $ is not a problem at all then, the shell won't do anything weird with it. Your [ -f ... ] is problematic though, since * returns more than one file sometimes, which would be a syntax error. So, how about this?

Code:

#!/bin/ksh

genfile="$1"

while IFS=", " read filedir name
do
        echo "filedir [$filedir] name [$name]"
        set -- "${name}"* # Overwrites $1, $2, etc with file1, file2, ...
        if [ -f "$1" ]
        then
                echo "Found file $1"
        fi
done < masterfile

No, it is not. A "variable" is something like "$name" or "${name}", where "name" is a name you choose. "$xyz" might be a variable, but "v$xyz" is a literal "v" followed by the content of a variable with the name "xyz".

OK, but is the "$xyz" in this filename now meant literally or should that signify the variable portion of the filename which you want to transport via the variable? If the first is the case a simple quotation will suffice, because this is what it is for - inside of (single) quotes the "$" loses its special meaning to the shell and reverts back to a normal character:

Code:

ls -l v$xyz_xxxx_xxxx.txt

will first replace "$xyz_xxxx_xxxx" with the content of a variable named "xyz_xxxx_xxxx" (which most probably will not exist so that it evaluates to "", the empty string) and therefore search for a file named "v.txt". Whereas:

Code:

ls -l 'v$xyz_xxxx_xxxx.txt'

(notice the single quotes) will look for a file named exactly this: "v$xyz_xxxx_xxxx.txt".

If "$xyz" is a variable and you want to use that as the changing part of an otherwise fixed filename then this code snippet is for you:

Code:

xyz="some$thing"
ls -l "v${xyz}_xxxx_xxxx.txt"

This will search for a file name "vsome$thing_xxxx_xxxx.txt". Because of the braces around "xyz" the shell knows that only this (and not the rest of the string) is the name of the variable. Notice the double quotes: you should habitually double-quote every use of any variable - ever and always! (There are only very few exceptions to this rule of thumb.) This way blanks (yes, filename can contain blanks) will not break your script.

I hope this helps.

bakunin

thanks everybody, it works. I am not sure what i did wrong yesterday. below is my dummy code

Code:

$1 is 'v$abc', filename is v$abc_xxxxx_20180120.txt, the value in filename is v$abc_xxxxx

Code:

#!/bin/ksh
feed=$1
echo $feed
while read filename
do
if [ -f ${filename}* ]; then
mv ${1}* ${1}_newname
fi
done<filename|grep ${1}