Here I have tried to find out the number of commas in a row. Notice that "n" numbers of names have been stored as a string being separated by n-1 numbers of commas in between. So each string is having n (n-1 +1) names. Any string having at least one name without any comma cannot even skip the logic.

Listing 2

select id,sum(length(name)-length(translate(name,' ,',' '))+1)NAME 
from multiple_id group by id;

Figure3