Re: Need bash script Help
- From: "Martin Slack" <m.g.slack@xxxxxxxxxxxx>
- To: <programmingblind@xxxxxxxxxxxxx>
- Date: Sun, 25 Jul 2010 08:50:24 +0100
Hi Don,
I may not be understanding what you want to achieve, but ignoring the
logic, I can get, using the function below without the use of eval or
multiple quoting:
$ function show
{
echo $1
_var="$2"
OFS="--"
_total="$1"$OFS"$2"
}
$ show "one two three" four
one two three
$ echo $_var
four
$ echo $_total
one two three--four
Martin
----- Original Message -----
From: "Donald Marang" <donald.marang@xxxxxxxxx>
To: <programmingblind@xxxxxxxxxxxxx>
Sent: Sunday, July 25, 2010 12:31 AM
Subject: Need bash script Help
I hope there is someone here with a little experience with functions in a
bash script. I wrote a function that will append a filename to a list of
filenames. The first argument passed is the name of the variable for a list
of filenames. I fashoned this method based on a bash tutorial on how to
return values from a function. that use this method and the 'eval'
statements within the if statement. Since bash functions can not returna
string, passing just the name seems to be a poor man's way to call by
reference. The second argument is a filename that will be appended to the
lisst.
I think everything is in order except the line below:
eval _ResultString="'$1'"
I need _ResultString to contain the actual filenames in the variable name in
the variable whose name is the first argument. For some reason I can not
get this statement correct.
Can anybody know how this assignment statement should look?
Here's the function:
AddFilenameToString() {
# $1 is the variable name of the sttring of filenames
# $2 is the new filename to Add
local _ResultName="$1"
local _ResultString=""
eval _ResultString="'$1'"
echo "variable name = $_ResultName"
echo "value = $_ResultString"
if [[ -z "$_ResultString" ]]; then
echo "create new string"
eval $_ResultName="'$2'"
else
echo "append to string"
eval $_ResultName="'${_ResultString}${IFS}$2'"
fi
}
Don Marang
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