Writing a (slightly) simple compilation script

[Brando569]

i still get the same thing whether i add the spaces in or not:

compile: 47: [[: not found
compile: 76: [[: not found

#!/bin/bash
clear
echo "version of keytouch? "
#read keytouch_version
keytouch_version="2.4.1"

echo "version of keytouch editor? "
#read editor_version
editor_version="3.2.0-beta"

echo "extract tarballs?"
read decision
echo ""

if [[ "$decision" == y]]; then

	tar xvf keytouch-$keytouch_version.tar.gz
	tar xvf	keytouch-editor-$editor_version.tar.gz
	echo ""

	#compiles and installs keytouch
	cd keytouch-$keytouch_version
	./configure
	make -s
	sudo make -s install
	echo ""

	cd keytouch-config
	./configure
	make -s
	sudo make -s install
	echo ""

	cd ../keytouch-keyboard
	./configure
	make -s
	sudo make -s install
	echo ""

	#compiles and installs the keytouch editor
	cd ../../keytouch-editor-$editor_version
	./configure
	make -s
	sudo make -s install
	echo ""

fi

if [[ "$decision" == n]]; then
	#compiles and installs keytouch
	cd 'keytouch-$keytouch_version'
	./configure
	make -s
	sudo make -s install
	echo ""

	cd keytouch-config
	./configure
	make -s
	sudo make -s install
	echo ""

	cd ../keytouch-keyboard
	./configure
	make -s
	sudo make -s install
	echo ""


	#compiles and installs the keytouch editor
	cd ../../keytouch-editor-$editor_version
	./configure
	make -s
	sudo make -s install
	echo ""
fi

for the other thing i want to be able to pass "own" the path in the same line, pretty much how you would do with any other command that requires a path, such as chown [user][path] or chgrp [group] [path]

[mssever]

I ran your code as posted, except that I put echo before all your compilation stuff since I don't have those files on my machine. With the spaces missing (as in your post--you still haven't added them as required), I get the following output:

~:$ ./test.sh 
version of keytouch? 
version of keytouch editor? 
extract tarballs?
y

./test.sh: line 15: syntax error in conditional expression: unexpected token `;'
./test.sh: line 15: syntax error near `;'
./test.sh: line 15: `if [[ "$decision" == y]]; then'

When I add a space where required (hint: after the y in the error above), it works fine.

Your error message looks like you aren't running bash. Are you sure you're running the code as you posted? You aren't doing something silly like

sh your_script

are you? That would produce the error you're getting.

If you add in the spaces I've mentioned multiple times, your script works. One suggestion, though. Why check for $decision == y then later for $decision == n? Why not use an else for the second case? It would make your code clearer.

for the other thing i want to be able to pass "own" the path in the same line, pretty much how you would do with any other command that requires a path, such as chown [user][path] or chgrp [group] [path]

See my previous post for the answer to this question.

[Brando569]

Your error message looks like you aren't running bash. Are you sure you're running the code as you posted? You aren't doing something silly like

sh your_script

are you? That would produce the error you're getting.

guilty as charged... i changed the permissions on it and executed it and it worked perfectly. thanks for all your help!

If you add in the spaces I've mentioned multiple times, your script works. One suggestion, though. Why check for $decision == y then later for $decision == n? Why not use an else for the second case? It would make your code clearer.[/QUOTE]

i tried to do that but then i changed it for some reason

[mssever]

guilty as charged... i changed the permissions on it and executed it and it worked perfectly. thanks for all your help!

If you had called it as bash your_script, it would have worked. Basically, you can forget that sh exists (except for a few specific situations) and just use bash.

[Brando569]

another lesson learned. ive always wondered this, whats the difference between running a script with sh and ./ ?

edit: the other thing that i wanted works.... mostly. for some reason chgrp can red the variable but chown cant

path="$1"; sudo chown -vR bran $path; sudo chgrp -vR bran $path

[mssever]

another lesson learned. ive always wondered this, whats the difference between running a script with sh and ./ ?

When you run your script directly (such as ./), the kernel examines the shebang line to find an interpreter. It then calls the interpreter named with the path to your script as its argument. When you explicitly call an interpreter, the interpreter simply evaluates the file as it sees fit. Shebang lines are comments, so they get ignored.

edit: the other thing that i wanted works.... mostly. for some reason chgrp can red the variable but chown cant

path="$1"; sudo chown -vR bran $path; sudo chgrp -vR bran $path

I'm not sure I understand what you mean when you say that chgrp can't read the variable. However, be sure to out all variable references in double quotes. Otherwise, if someone passes in a path that contains spaces, your script will break.

path="$1"
sudo chown -vR bran:bran "$path" # you can set both owner and group in a single command

[Brando569]

thanks for the explanation and i never knew you could do both user and group in one command, ive been doing them seperate for years! it works now and what i meant before was that when i would run the script it would say that chown was missing an arugment after bran (the path) yet chgrp wouldnt complain of anything.