Shell script input

Question

#!/bin/sh 

if test -n $1  
then  
  echo "Some input entered"  
  echo $1  
else  
  echo "no input entered"  
fi

The above code is supposed to say "no input entered" if I dont pass an argument to the shell script. the echo $1 shows a blank line when I dont pass any arguments. its saying "some input entered" even without any arguments.

Are you deliberately trying to maintain bourne shell compatibility? Can't you use #!/bin/bash? And finally, if you're using bash, forget test and start using if [[ cond ]] ; then form of syntax. Good luck. — shellter
– shellter, Commented Aug 3, 2011 at 0:50

evil otto · Accepted Answer · 2011-08-03 00:46:59Z

5

put quotes around your $1. Without them, the $1 just vanishes and test confusingly reports "nothing" to be not empty.

if test -n "$1" 
then
    ....

answered Aug 3, 2011 at 0:46

evil otto

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Comments

Chris Gregg · Accepted Answer · 2011-08-03 00:48:57Z

1

Skip test and put quotes around the $1:

#!/bin/sh 

if [ -n "$1" ]
then  
  echo "Some input entered"  
  echo $1  
else  
  echo "no input entered"  
fi

answered Aug 3, 2011 at 0:48

Chris Gregg

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2 Comments

evil otto Over a year ago

[ is just a synonym for test

Chris Gregg Over a year ago

@evil otto - I did not know that, but now that I look at the test manpage, I see that you are correct! :) @dasman should give you the check, as you had the answer first.

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