Arguments to change variable values in Bash script - bash

Question

i have this script in bash:
#!/bin/bash
dir="/home/dortiz/Prueba"
for i in $dir/*
do
cat $i | awk '{print $1" " $2" " $3" " $4"\n " $5}' | \
awk '/gi/{print ">" $0; getline; print}' | \
awk '$3>20.00 {print $0; getline; print;}' \
> "${i}.outsel"
done
cd /home/dortiz/Prueba
mv *.outsel /home/dortiz/Prueba2
and i would like to set an argument to change the value after ""awk '$3>"" in an easy way from my main program that will call this script.
i have read something about getopts but i dont uderstand it at all
Thanks a lot in advance

Answer 1

The simplest way is to just pass an argument to your script:
yourscript.sh 20.0
Then in your script
#!/bin/bash
value=$1 # store the value passed in as the first parameter.
dir="/home/dortiz/Prueba"
for i in $dir/*; do
awk '{print $1" " $2" " $3" " $4"\n " $5}' "$i" |
awk '/gi/{print ">" $0; getline; print}' |
awk -v val="$value" '$3>val {print $0; getline; print;}' > "${i}.outsel"
# ^^^^^^^^^^^^^^^
done
...
and the cat|awk|awk|awk pipeline can probably be written like this:
awk -v val="$value" '
$3 > val {
prefix = /gi/ ? ">" : ""
print prefix $1 " " $2" " $3" " $4"\n " $5
}
' "$i" > "$i.outsel"