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Shell Programming and Scripting

Unix commands in one line

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# 8  
Old 11-03-2008
Code:
echo "file$([[ ! -f ./test_file_file ]] && echo " not") there"

# 9  
Old 11-03-2008
If I was going to use && I would use || as well rather than doing the test twice. Also no need to assign $? to a variable, as that is the result that && and || calculate their logic on anyway:

Code:
ls ./test_file_file && echo "file there" || echo "file not there"

# 10  
Old 11-03-2008
Quote:
Originally Posted by Annihilannic
If I was going to use && I would use || as well rather than doing the test twice. Also no need to assign $? to a variable, as that is the result that && and || calculate their logic on anyway:

Code:
ls ./test_file_file && echo "file there" || echo "file not there"


I wouldn't use ls; it is unnecessary for testing the file's existence. I'd use the shell's built-in test:

Code:
[ -f ./test_file_file ] && echo "file there" || echo "file not there"

# 11  
Old 11-03-2008
I agree, unless you wanted the output of the ls command to be displayed as well for some reason.
# 12  
Old 11-04-2008
Quote:
Originally Posted by Annihilannic
I agree, unless you wanted the output of the ls command to be displayed as well for some reason.

In which case it would be faster to use echo or printf to display the name of the file; there's no need to use an external command to do it.
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