5

Please tell me why I couldn't print variable value.

# chmod 777 chkscript.sh
# ./chkscript.sh

chkscript.sh file content

variable = "This is variable"

echo "$variable"

echo "Hello World "

Output :

# ./chkscript.sh
./chkscript.sh: line 5: variable: command not found

Hello World
#

P.S : And sometimes

 variable1 = "/home/files" --- which is location if I try to print nothing gets printed.
 echo "$variable"
10

In shell, spaces are not allowed on either side of the = in a variable assignment.

Try this instead:

variable="This is a variable"

If you leave a space before the =, the shell parses the token before it as a command or function name, which is why you see the "command not found" message.

| improve this answer | |
  • variable="This is variable" echo"$variable" changed like this but no use.Thanks for reply. – sunleo Jul 2 '14 at 11:53
  • 1
    You still need the space after echo - if you omit it, the shell expands what's in $variable and then tries to find the command that is called (in this example) echoThis with the arguments is variable. Almost certainly NOT what you want! – D_Bye Jul 2 '14 at 11:55
  • 1
    Thanks for the insight on that. I am still learning shell scripting as I go :) – ryekayo Jul 2 '14 at 11:56
  • Finally fixed this Thanks you.It is really hard to fix space related thing here.Thanks for all your replys. – sunleo Jul 2 '14 at 11:57
-1

I had a similar problem, and I did this:

#!/bin/sh
my_chars='This is test' ;
echo $my_chars
~

Now

$ ./test_chars.sh
This is test

This works

| improve this answer | |
  • This doesn't really explain what was wrong. – X Tian Aug 23 '16 at 13:37

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