I have a text which looks like -
Application.||dates:[2022-11-12]|models:[MODEL1]|count:1|ids:2320
Application.||dates:[2022-11-12]|models:[MODEL1]|count:5|ids:2320
I want the number from the count:1
columns so 1
and i wish to store these numbers in an array.
nums=($(echo -n "$grepResult" | awk -F ':' '{ print $4 }' | awk -F '|' '{ print $1 }'))
this seems very repetitive and not very efficient, any ideas how to simplify this ?
You can use awk once, set the field separator to |
. Then loop all the fields and split on :
If the field starts with count
then print the second part of the splitted value.
This way the count:
part can occur anywhere in the string and can possibly print this multiple times.
nums=($(echo -n "$grepResult" | awk -F'|' '
{
for(i=1; i<=NF; i++) {
split($i, a, ":")
if (a[1] == "count") {
print a[2]
}
}
}
'))
for i in "${nums[@]}"
do
echo "$i"
done
Output
1
5
If you want to combine the both split values, you can use [|:]
as a character class and print field number 8 for a precise match as mentioned in the comments.
Note that it does not check if it starts with count:
nums=($(echo -n "$grepResult" | awk -F '[|:]' '{print $8}'))
With gnu awk
you can use a capture group to get a bit more precise match where on the left and right can be either the start/end of string or a pipe char. The 2nd group matches 1 or more digits:
nums=($(echo -n "$grepResult" | awk 'match($0, /(^|\|)count:([0-9]+)(\||$)/, a) {print a[2]}' ))