I assign command ls to my variable my_File but when I run it as $my_File it does not work. Can you please explain why?
#!/bin/bash
my_File='ls -f | grep -v '\/''
$my_File
4 Answers
The line you wrote defines a variable whose value is the string ls -f | grep -v /. When you use it unquoted, it expands to a list of words (the string is split at whitespace): ls, -f, |, grep, -v, /. The | character isn't special since the shell is splitting the string, not parsing it, so it becomes the second argument to the ls command.
You can't stuff a command into a string like this. To define a short name for a command, use an alias (if the command is complete) or a function (if the command has parameters and is anything more than passing some default arguments to a single command). In your case, an alias will do.
alias my_File='ls -f | grep -v /'
my_File
answered Mar 26, 2014 at 0:21
Gilles 'SO- stop being evil'
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$ my_File='ls -f | grep -v '\/''
$ $my_File
ls: cannot access |: No such file or directory
ls: cannot access grep: No such file or directory
[snip]
When interpreting $my_File, bash treats the characters in it as just characters. Thus, for one, the command line has a literal | character in it, not a pipe.
If you are trying to execute $my_File on the command line and have the pipes work, you need eval $my_File.
echo "${var='ls -f | grep -v "\/"'}" |sh
You certainly need an interpreter - though not necessarily eval.
. <<-HEREDOC /dev/stdin
$var
HEREDOC
echo "$var" | . /dev/stdin
There are a lot of ways to get there.
${0#-} -c "$var"
sh - c "$var"
variable = command
backquote character(`)
so like
var = pwd
echo $var = /home/user/foo
-
2That would assign the output of the command to the variable which is not what is being asked for here.Stéphane Chazelas– Stéphane Chazelas2017-01-20 10:45:19 +00:00Commented Jan 20, 2017 at 10:45
my_File=$(ls -f | grep -v '\/')or substitute the outer quotes by backquotes...alias?'ls -f | grep -v '\/''supposed to be?