unary operator expected

Shell Programming and Scripting
1

hi

i am trying to compare a value with value 50. but i am getting
"[: -lt: unary operator expected"

I am using

 if [ $j -lt 50 ]
then 
   echo "------------"

fi

please help

thanks in advance
Satya

2

If $j has no value at all, the expression will be invalid because it will vanish entirely, making it look like just [ -j 50 ]. Wrap $j in double-quotes and it won't vanish, becoming [ "" -lt 50 ] , which still isn't altogether sensible -- comparing a blank string to an integer -- but at least has correct syntax.

3

Make sure that $j expands to a value. Best is to set it explicitly, but you can also give it a default value (0 in this example):

if [ ${j:-0} -lt 50 ]