use conditional in bash script to check string argument

Question

I am trying to write my shell script thing.sh so that upon making it an executable and running it with the single letter ``A" like so:

$ ./thing.sh A 

I get the output

A  

If argument 1 is not A, I want the output

Not A  

Here is my code so far :

#!/bin/bash  if [ "$1" -eq "A"] then echo "A" else echo "Not A" fi  

which returns, no matter what I enter,

./thing.sh: line 3: [:missing `]' Not A 

I am trying what I hoped would check something with one or several letters and compare it against the letter A; could someone tell me what I am missing to get this to work? Thank you

Answer 1

What about the shorter :

#!/bin/bash  [[ $1 == A ]] && echo "A" || echo "not A" 

?

And a beginner version (identical logic) :

#!/bin/bash  if [[ $1 == A ]]; then     echo "A" else     echo "not A" fi 

Like Scott said, you have a syntax error (missing space).

explanations

• I use boolean logic here. [[ $1 == A ]] is executed, and then if its true, echo "A" is executed, and if it's false, echo "not A" is executed, See http://mywiki.wooledge.org/BashGuide/TestsAndConditionals
• [[ is a bash keyword similar to (but more powerful than) the [ command. See http://mywiki.wooledge.org/BashFAQ/031 and http://mywiki.wooledge.org/BashGuide/TestsAndConditionals Unless you're writing for POSIX sh, I recommend [[.

Answer 2

Change the first line to:

if [ "$1" == "A" ] 

The -eq operator is for integers. And as someone else mentioned, the space does matter before the ']'.

See here: http://tldp.org/LDP/abs/html/comparison-ops.html