how to compare a variable with a number?

pdklinux79 · 06-10-2008, 06:41 PM

the code is as below:

if [ $i -le $tnum ] ;
then
if [ $t -ne 0 ];
then
{array[$tnum]}=$t
tnum=$((tnum+1))
fi
else
{array[$tnum]}=$t
tnum=$((tnum+1))
fi

bash :
error :./omrprac1.sh: line 14: [: 0: unary operator expected

./omrprac1.sh: line 14: [: 0: unary operator expected

./omrprac1.sh: line 14: [: 0: unary operator expected

./omrprac1.sh: line 14: [: 0: unary operator expected

how can i compare a a variable t with number zero?

chrism01 · 06-10-2008, 06:50 PM

Tricky, there's only 11 lines there ....

Try it this way (inc indentations; use code tags)

Code:

#!/bin/bash
# this next line turns on comprehensive debugging
set -xv

# need to init vars
i=1
tnum=2

# Note: do not use ';' if the 'then' statement is on a separate line
if [[ $i -le $tnum ]]
then
    if [[ $t -ne 0 ]]
    then
        {array[$tnum]}=$t
        tnum=$((tnum+1))
    fi
else
    {array[$tnum]}=$t
    tnum=$((tnum+1))
fi

http://tldp.org/LDP/Bash-Beginners-G...tml/index.html
http://www.tldp.org/LDP/abs/html/

HTH

pdklinux79 · 06-10-2008, 07:02 PM

the below is the code:
#!/bin/bash

i=1
tnum=1

while read line
do
t=$(echo "$line" | awk '{print $3}')
if [ $i -ne $tnum ]
then
if [ $t -ne 0 ]
then
{array[$tnum]}=$t
echo ${array[$tnum]} #not showing the value
tnum=$((tnum+1))
fi
else
{array[$tnum]}=$t
echo ${array[$tnum]} #not showing the value
tnum=$((tnum+1))
fi
echo ${array1[$tnum]} #not showing the v

done < <(grep '^ID' /home/admin/cdat.txt)

please let me know why the array is not echoed when i run the program.. it is not taking at all...