How to count number of values in a row and store total count to array - shell

Question

I have a scenario where i want to get count of all values row by row and store it to dynamic array
Data in file :
"A","B","C","B"
"P","W","R","S"
"E","U","C","S"
"Y","F","C"
first row as : 4 -> values
second row as : 4 -> values
third row as : 4 -> values
fourth row as : 3 -> values
Expected Output :
store to array : array_list=(4,4,4,3)
written a script but not working
array_list=()
while read -r line
do
var_comma_count=`echo "$line" | tr -cd , | wc -c`
array_list=+($( var_comma_count))
done < demo.txt
when i print array it should give me all values : echo "{array_list[#]}"
Note :
The file might contain empty lines at last which should not be read
when i count file it gave me count : 5 , it should have ignored last line which is empty
where as when i use awk it give me proper count : awk '{print NF}' demo.txt -> 4
I know processing file using while loop is not a best practise , but any better solution will be appreciated

Answer 1

Perhaps this might be easier using awk, set the FS to a comma and check if the number of fields is larger than 0:
#!/bin/bash
array_list=($(awk -v FS=, 'NF>0 {print NF}' demo.txt))
echo "${array_list[#]}"
Output
4 4 4 3
The awk command explained:
awk -v FS=, ' # Start awk, set the Field Separator (FS) to a comma
NF>0 {print NF} # If the Number of Fields (NF) is greater than 0, print the NF
' demo.txt # Close awk and set demo.txt as the input file
Another option could be first matching the format of the whole line. If it matches, there is at least a single occurrence.
Then split the line on a comma.
array_list=($(awk '/^"[A-Z]"(,"[A-Z]")*$/{print(split($0,a,","));}' demo.txt))
echo "${array_list[#]}"
Output
4 4 4 3
The awk command explained:
awk '/^"[A-Z]"(,"[A-Z]")*$/{ # Regex pattern for the whole line, match a single char A-Z between " and optionally repeat preceded by a comma
print(split($0,a,",")); # Split the whole line `$0` on a comma and print the number of parts
}
' demo.txt