On Thu, 25 Jan 2001 at 10:21am (-0800), Brad Doster wrote:
> 1) I have a variable with a list of variable names as its content, e.g.
> 'varlist=$var1 $var2 $var3'. I want to manipulate $varlist such that it
> ='var1 var2 var3', i.e get rid of the '$'s in front of each variable name.
> A 'sed' example that I've tried is:
>
> varlist=`echo "$varlist" | sed 's/$//'`
>
> I've tried every variation I can think of with and without "s around
> '$varlist', using single and double quotes for the sed command, and escaping
> the $ with \. In short, nothing works -- it always comes back with '$var1
> $var2 $var3'.
Sounds like your doing the right thing... I don't know why it whoudl not be
working.
$ varlist='$var1 $var2 $var3'
$ echo $varlist
$var1 $var2 $var3
$ echo $varlist | sed 's/\$//g'
var1 var2 var3
$
Not using the g on the end of the sed command would cause it to only replace
the first $ but I would still have expected it to do /something/.
$ echo $varlist | sed 's/\$//'
var1 $var2 $var3
$
> I did find that 'varlist=`echo $varlist | awk -F "$" '{print $2}'` does
> work, but it seems there ought to be a way to do it with sed. Any idea what
> I'm missing?
>
> 2) Given two variables, say var1="a" and var2="b", I want to create var3
> such that it is "a<newline>b", i.e 'echo "$var3"' produces:
>
> a
> b
>
> Combinations like 'var3=$var1\n$var2' end up as 'anb', i.e. the '\n' is not
> treated as a <newline> character. Again, I've tried it with a variety of
> quoting patterns, etc., all to no avail. So again...
>
This parts a bit trickier I think. Even if you get the linefeed in there
bash is not inclined to show it to you. It treats a linefeed as a word
seperator so a space is a tab is a linefeed. IFS lets us define what is to
be concidered as seperators so we can get round this.
$ var1=a
$ var2=b
$ var3="$var1
> $var2"
$ echo $var3
a b
$ IFS=' '
$ echo $var3
a
b
$
I don't know if that's a workable 'solution' in your case.
In the example I wrote the assignment on two seperate lines but you could
also do something like...
var3="$(echo $var1; echo $var2)"
... or..
var3="$(echo -e $var1\\n$var2)"
... you just need something else to provide the line feed becuase bash won't
evaluate \n directly.
M.
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