[bash-script] Problem by "if [ -e $NAME ]; then" ...

thomas2004ch · 03-02-2012, 05:25 AM

Hi,

My script looks as follow:

Code:

#!/bin/sh

NAME=$(find ./out -type f -name "test*")

echo Name=$NAME

if [ -e $NAME ]; then
  echo "file exist"
fi

There is no file under ./out but it returns that there is file test* exist as follwo. Why?

Code:

Name=
file exist

acid_kewpie · 03-02-2012, 05:49 AM

well this is clearly nothing to do with find itself.

You can reduce it to this:

Code:

if [ -e ]
then
   echo "file exist"
fi

and that "works", which I don't understand. I suppose you could say that "nothing" does exist, but it's a bit philosophical really!

don't use [ ], under bash, use [[ ]] instead and you'll have a better time in general as [ is the test binary, whilst [[ is a bash built in, so it *KNOWS* about the NAME variable you have there, like a normal perl / c program would, but [ / test just sees if after string substitutiong have occurred. Alternatively, if you do use [, ALWAYS put your variables in double quote marks.

thomas2004ch · 03-02-2012, 06:46 AM

I reduce the code as you wrote but it doesn't work.

Besides I am not sure if we can use if [ -e ] without an operant.

whizje · 03-02-2012, 06:58 AM

An alternative maybe

Code:

bash-4.1$ find ./out -type f -name "test*" -printf 'name=%p\n' -exec echo "file exist" \;
name=./out/test
file exist
bash-4.1$ rm out/test
bash-4.1$ find ./out -type f -name "test*" -printf 'name=%p\n' -exec echo "file exist" \;
bash-4.1$ >out/test
bash-4.1$ >out/test1
bash-4.1$ find ./out -type f -name "test*" -printf 'name=%p\n' -exec echo "file exist" \;
name=./out/test
file exist
name=./out/test1
file exist
bash-4.1$ find ./out -type f -name "test*" -printf 'name=%p\nfile exist\n' 
name=./out/test
file exist
name=./out/test1
file exist
bash-4.1$

thomas2004ch · 03-02-2012, 07:04 AM

Now I change the

Code:

...
if [ -e $NAME ]; then
...

to

Code:

...
if [[ -z $NAME ]];  then
...

And this works.

thomas2004ch · 03-02-2012, 07:05 AM

Quote:

Originally Posted by whizje

An alternative maybe

Code:

bash-4.1$ find ./out -type f -name "test*" -printf 'name=%p\n' -exec echo "file exist" \;
name=./out/test
file exist
bash-4.1$ rm out/test
bash-4.1$ find ./out -type f -name "test*" -printf 'name=%p\n' -exec echo "file exist" \;
bash-4.1$ >out/test
bash-4.1$ >out/test1
bash-4.1$ find ./out -type f -name "test*" -printf 'name=%p\n' -exec echo "file exist" \;
name=./out/test
file exist
name=./out/test1
file exist
bash-4.1$ find ./out -type f -name "test*" -printf 'name=%p\nfile exist\n' 
name=./out/test
file exist
name=./out/test1
file exist
bash-4.1$

Truely to say I don't understand what you mean here.

whizje · 03-02-2012, 07:09 AM

You could use only find when a name with test is found printf is executed

Code:

-printf 'name=%p\nfile exist\n'

name= is just printed
%p is the name of the file found
\n is a newline
file exist is also just printed
\n another newline

acid_kewpie · 03-02-2012, 07:11 AM

Quote:

Originally Posted by thomas2004ch

Now I change the

Code:

...
if [ -e $NAME ]; then
...

to

Code:

...
if [[ -z $NAME ]];  then
...

And this works.

Yes, that's absolutely what I'd recommend.

The bit *I* didn't understand is how you CAN use -e without an operant, but you can, it doesn't error, and that was actually *exactly* what you were doing, as [ doesn't evaluate it's contents until after $NAME is already long gone.