Jake Wilson Jake Wilson - 4 months ago 6x
Bash Question

Bash script variable declaration - command not found

This seems like such a simple question I'm embarrassed to ask it:


STR = "Hello World"
echo $STR

when I run sh test.sh I get this:

test.sh: line 2: STR: command not found

What am I doing wrong? I look at extremely basic/beginners bash scripting tutorials online and this is how they say to declare variables... So I'm not sure what I'm doing wrong.

I'm on Ubuntu Server 9.10. And yes, bash is located at


You cannot have spaces around your '=' sign.

When you write:

STR = "foo"

bash tries to run a command named STR with 2 arguments (the strings '=' and 'foo')

When you write:

STR =foo

bash tries to run a command named STR with 1 argument (the string '=foo')

When you write:

STR= foo

bash tries to run the command foo with STR set to the empty string in its environment.

I'm not sure if this helps to clarify or if it is mere obfuscation, but note that:

  1. the first command is exactly equivalent to: STR "=" "foo",
  2. the second is the same as STR "=foo",
  3. and the last is equivalent to STR="" foo.

The relevant section of the sh language spec, section 2.9.1 states:

A "simple command" is a sequence of optional variable assignments and redirections, in any sequence, optionally followed by words and redirections, terminated by a control operator.

In that context, a word is the command that bash is going to run. Any string containing = (in any position other than at the beginning of the string) which is not a redirection is a variable assignment, while any string that is not a redirection and does not contain = is a command. In STR = "foo", STR is not a variable assignment.