Processing text with multiple delims in awk - bash

Question

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 ?

Answer 1

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]}' ))

Answer 2

Try sed
nums=($(sed 's/.*count://;s/|.*//' <<< "$grepResult"))
Explanation:
There are two sed commands separated with ; symbol.
First command 's/.*count://' remove all characters till 'count:' including it.
Second command 's/|.*//' remove all characters starting from '|' including it.
Command order is important here.