UNIX for Dummies Questions & Answers

Hi,
No need to say I'm new to unix shell scripting.
I have a very simple script that goes this way:

for datos in `ls -rt $UNXLOG/26-Jan*`
do
export arch=`echo $datos |cut -d, -f1`
if [ `grep -c INACTIVO ${arch}` -eq 0 ]
then
export linea1=`grep Debut ${arch}`
export horatot=`echo $linea1 |cut -d' ' -f5`
export hora=`echo $horatot |cut -c1-2`

if [ ${hora} -le 19 ]
then
echo "Listando log - ${arch} ...\n" >> $UNXLOG/qq.log
more ${arch} >>$UNXLOG/qq.log
echo "--------------------------------------------------------------------------------" >>$UNXLOG/qq.log
fi
fi
done

When I execute the script, I get the error:

test: argument expected showing me the line in error is "if [ ${hora} -le 19 ]"

Any ideas????
Thks in advance!

should be :
if [ `echo $hora` -le 19 ]

Tried it out but still doesn't work.

Does it show the same error?

It would help if you put in some sample of the file/text you are cutting - it's possible you are ending up with nothing in that variable.

The problem is that hora has no value at all. Put a
echo hora = $hora
in front of the if statement to see that. This means that the if statement is just:

Code:

if [       -le 19 ]

One solution is to expand hora with a default value:

if [ ${hora:-1} -le 19 ]

Now if hora is unset or set to null, the if statement will see 1.

Thks!!!! I was able to solve it adding the default value you suggested!!!!!