Hi all
I have a question related with variables. Let's assume that I have one .txt file in my /tmp directory. Look here:

#!/bin/sh
var='/tmp/*.txt'
echo var

I'm getting
/tmp/file.txt

That's ok

Now I want the file name to be displayed in this sitaution to :
echo "file=>$var"

It shows
file=>/tmp/*.txt

But I want it to be shown as
file=>/tmp/file.txt

How can I solve it?
Thanks ...

Throw an ls in there. You want something like ls *.txt. I'll let you figure out the rest.

You can try printf in place of echo:

Thank you colucix for reply
I can't put it afte <<EOF

It doesn't understand it
I have to put it between <<EOF EOF

When I put $var, it understands it as /tmp/*.txt
When I put printf or echo, it doesn't understand it as LINUX command. Who to be??

Thanks ..
PLease show me an example which you 'll put it inside <<EOF EOF

Thanks ..

First thing:

You probably mean "echo $var", and not "echo var", which would output the string "var" textually, and not the contents of the $var variable.

When trying to expand the wild card with the string "file=>" in front of $var it won't find matches, since it becomes part of the path breaking any meaning it might have. A way to proceed would be to use ls as someone else said, but beware that you need to use the proper syntax to avoid color codes which could screw your script at any given moment. You probably will also have problems with spaces in the middle of the names, depending on what are you doing.

There are many better ways to overcome this, for example, you can add the file=> string at a later stage, after the wildcard expansion has taken place, like this:

EDITED: By the way, if you don't mind having all the file names on a single line you could as well use separate echos.

Thanks for response

I solved it using
<<EOF

file=>'$(printf $var)'

EOF