Too many arguments error in shell script - shell

Question

I am trying a simple shell script like the following:
#!/bin/bash
up_cap=$( cat result.txt | cut -d ":" -f 6,7 | sort -n | cut -d " " -f 2 | sort -n)
down_cap=$( cat result.txt | cut -d : -f 6,7 | sort -n | cut -d " " -f 6| sort -n)
for value in "${down_cap[#]}";do
if [ $value > 80000 ]; then
cat result.txt | grep -B 1 "$value"
fi
done
echo " All done, exiting"
when I execute the above script as ./script.sh, I get the error:
./script.sh: line 5: [: too many arguments
All done, exiting
I have googled enough, and still not able to rectify this.

Answer 1

You want
if [ "$value" -gt 80000 ]; then
You use -gt for checking if A is bigger than B, not >. The quotation marks I merely added to prevent the script from failing in case $value is empty.

Answer 2

Try to declare variable $value explicitly:
declare -i value
So, with the dominikh's and mine additions the code should look like this:
#!/bin/bash
up_cap=$( cat result.txt | cut -d ":" -f 6,7 | sort -n | cut -d " " -f 2 | sort -n)
down_cap=$( cat result.txt | cut -d : -f 6,7 | sort -n | cut -d " " -f 6| sort -n)
for value in "${down_cap[#]}";do
declare -i value
if [ $value -gt 80000 ]; then
cat result.txt | grep -B 1 "$value"
fi
done
echo " All done, exiting"