Percentage Queries

[bangemd]

I have tried the code in my database and table and it works!

Pls check your code for the table name as mine is Table1.
Others are the same as yours.

Hey thank you it work flawlessly. I was trying to learn so I didn't used the fruit example. I create a new database and the field is just confusing trying to match the fruit example

thanks a lot guys

[bangemd]

one last question relate to the problem how would I count the blank cell for example if one cell people haven't check out if it is a good or bad apple yet how would i incorporate in into the sql

[thhui]

The new revised code for your new question of including null status is as follows:-

SELECT fruit, 
Count([fruit])/(Select Count([fruit]) from Table1) AS [% of Fruit], 
Sum(iif([Status]="Good",1,0))/Count([fruit]) AS [% of Good], 
Sum(iif([Status]="Bad",1,0))/Count([fruit]) AS [% of Bad], 
Sum(iif(isnull([Status]),1,0))/Count([fruit]) AS [% of NULL] 
FROM Table1 
GROUP BY fruit;

[bangemd]

Thank you everyone I learned so much form you guy