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Shell Programming and Scripting

what's wrong with my match pattern

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Top Forums Shell Programming and Scripting what's wrong with my match pattern
# 1  
Old 08-27-2012
what's wrong with my match pattern
Code:
# [[ -f "/boot/vmlinuz*" ]] && echo "ok"
# ls /boot/vmlinuz-2.6.32-279.el6.x86_64 
/boot/vmlinuz-2.6.32-279.el6.x86_64

it doesn't return ok , any one help?
# 2  
Old 08-27-2012
In the [[...]] construct, the word following -f has to be a file-name path and not a pattern for a file-name path. -f will treat your provided pattern as a literal file-name and the other statement will not get executed.
An alternative to get what you want could be:
Code:
a=$(print /boot/vmlinuz*)
[[ $a != "/boot/vmlinuz*" && -f $a ]] && echo OK

I am assuming only one file in the directory with that pattern or none.

If multiple files or directories with that pattern may exist and if you are not averse to using a simple loop, you may try:
Code:
ls -d /boot/vmlinuz*|while read
do
 [[ -f $REPLY ]] && echo "$REPLY" is OK
done

Last edited by elixir_sinari; 08-27-2012 at 05:38 AM..
# 3  
Old 08-27-2012
Quote:
Originally Posted by elixir_sinari
Code:
a=/boot/vmlinuz*
[[ $a != "/boot/vmlinuz*" && -f $a ]] && echo OK

Hi elixir_sinari,

I think your first/previous code is correct...

Code:
a=/boot/vmlinuz*
echo $a
/boot/vmlinuz-2.6.32-279.el6.x86_64
[[ $a == /boot/vmlinuz* ]] && echo OK
OK

# 4  
Old 08-27-2012
Check now. I'd made a mistake. When a variable assignment such as a=/boot/vmlinuz* is made, the value stored in a will be /boot/vmlinuz* and nothing else. When that variable is evaluated, the expansion of * is done depending on whether you quote it or not.
# 5  
Old 08-27-2012
Quote:
Originally Posted by elixir_sinari
Check now. I'd made a mistake. When a variable assignment such as a=/boot/vmlinuz* is made, the value stored in a will be /boot/vmlinuz* and nothing else. When that variable is evaluated, the expansion of * is done depending on whether you quote it or not.
Code:
it still not ok 
# kernel="/boot/vmlinuz*"
# echo $kernel
/boot/vmlinuz-2.6.32-279.el6.x86_64
[[ -f $kernel ]] && echo "ok" 
#

# 6  
Old 08-27-2012
Check my first post in this thread.
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