[SOLVED] bash problem. syntax error?

[spacesearcher]

ok so i am trying to learn to write bash scripts. I found this page that is helping: http://www.ibm.com/developerworks/library/l-bash2.html
im trying to do the String comparison caveats section. so i write this script


#!/bin/bash

if [ "$myvar" = "foo bar oni" ]
then
echo "yes"
fi

then save it as myscript.sh
in terminal i cd to the directory and do:


joe@joe-laptop:~/Computer practice$ chmod 755 myscript.sh
joe@joe-laptop:~/Computer practice$ myvar="foo bar oni"
joe@joe-laptop:~/Computer practice$ ./myscript.sh
joe@joe-laptop:~/Computer practice$ echo $myvar
foo bar oni
joe@joe-laptop:~/Computer practice$

it never echos yes like it should. >.< this is driving me crazy. all examples have worked but this one hasn't. Am I just not understanding something?

[ghostdog74]

use case


case "$myvar" in
"foo bar oni" ) echo "yes";;
* ) echo "no";;
esac

[spacesearcher]

thank you!
I think I will have a good look at the link in your sig to bash scripting. it seems like more what im after.

[WW]

You need to export the variable for it to be visible in the script, or execute the script using the bash builtin command source (or define the variable in the script itself). Here is myscript.sh with an extra echo statement:

myscript.sh


#!/bin/bash
echo "myvar is '$myvar'"
if [ "$myvar" = "foo bar oni" ]
then
echo "yes"
fi


Here it is in use:


$ ./myscript.sh
myvar is ''
$ myvar="foo bar oni"
$ ./myscript.sh
myvar is ''
$ source myscript.sh
myvar is 'foo bar oni'
yes
$ export myvar
$ ./myscript.sh
myvar is 'foo bar oni'
yes
$