Bash while loop behaves differently dealing with multiple test conditions

Question

I wonder if someone could explain why a while loop treats multipe test conditions differently to an if loop. I have 2 tests that i verified came out as True and False:

Bash$ test ! -n "$(find . -maxdepth 1 -name '*.xml' -print -quit)"; echo $?
0
Bash$ test ! -e "unsentData.tmp"; echo $?
1
Bash$

When I ANDed these 2 tests into an if statement I got an aggregate of False as expected:

Bash$ if [ ! -n "$(find . -maxdepth 1 -name '*.xml' -print -quit)" ] && [ ! -e "unsentData.tmp" ]; then echo "True"; else echo "False"; fi
False
Bash$

Now when I put the 2 tests into a while loop I expected a sleep until both conditions were met but instead I got immediately a true result.

Bash$ while [ ! -n "$(find . -maxdepth 1 -name '*.xml' -print -quit)" ] && [ ! -e "unsentData.tmp" ]; do sleep 1; done; echo -e "All files Exist\n$(ls /opt/pcf/mfe/unsentXmlToTSM/xmlConnection0_TSM/)"
All files Exist 
unsentData.tmp
Bash$

What am I missing here? I simply want to write something that waits until the 2 conditions are met before it breaks out of the loop

A

Mithrandir · Accepted Answer · 2012-02-16 13:54:06Z

3

Your assumption is worng i think. While does execute the code between do and done while (as long as) the condition holds true. Your conditions combined evaluate to false, as seen in the output of your if-statement. Thus the body of the while loop never gets executed. Try:

while ! ( [ ! -n "$(find . -maxdepth 1 -name '*.xml' -print -quit)" ] && 
          [ ! -e "unsentData.tmp" ] )  
do 
    sleep 1
done 
echo -e "All files Exist\n$(ls /opt/pcf/mfe/unsentXmlToTSM/xmlConnection0_TSM/)"

edited Feb 16, 2012 at 13:54

answered Feb 16, 2012 at 12:43

Mithrandir

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2 Comments

amadain Over a year ago

That syntax does not seem to be correct. I got the error: -bash: syntax error near unexpected token `('. But I see your point

amadain Over a year ago

It still did not work. All the negations make it very confusing. Gordons suggestion of using an until loop set me in the correct track. Thanks

Gordon Davisson · Accepted Answer · 2012-02-16 18:15:56Z

1

A while loop executes as long as ("while") its condition is true; it sounds like you want to run the loop until its condition is true. bash has an until loop which does exactly this:

until [ ! -n "$(find . -maxdepth 1 -name '*.xml' -print -quit)" ] && [ ! -e "unsentData.tmp" ]; do
    sleep 1
done
echo -e "All files Exist\n$(ls /opt/pcf/mfe/unsentXmlToTSM/xmlConnection0_TSM/)"

Or you can just negate the condition (i.e. use "while there are files left, do..." rather than "until all files are done, do..."). In this case that just means removing the negations of the individual conditions and switching the and to an or:

while [ -n "$(find . -maxdepth 1 -name '*.xml' -print -quit)" ] || [ -e "unsentData.tmp" ]; do
    sleep 1
done
echo -e "All files Exist\n$(ls /opt/pcf/mfe/unsentXmlToTSM/xmlConnection0_TSM/)"

answered Feb 16, 2012 at 18:15

Gordon Davisson

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1 Comment

amadain Over a year ago

The until loop worked when I removed the negation. It was the direction I was looking for but overlooked. Thanks. I wanted to wait until both xml files and unsentData.tmp were present so no negation in either of the ANDed test conditions worked

kev · Accepted Answer · 2012-02-16 12:48:01Z

1

while [ ! -n "$(find . -maxdepth 1 -name '*.xml' -print -quit)" ] && [ ! -e "unsentData.tmp" ]
    do sleep 1
done

==>

while false
    do sleep 1
done

so the do sleep 1 didn't run at all.

answered Feb 16, 2012 at 12:48

kev

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