MySql - if there is no record, insert multiple rows - php

Question

Before asking this question, I already search a lot of entries on Google and StockOverflow. Nothing can fulfil my question.
There are two tables - group_sale_bonuses and members. I want to check is already there records with product_id "1" in the group_sale_bonuses.
If not, I want to insert all records from members table into group_sale_bonuses with product_id "1".
My overall requirement is as follow:
IF ((Select count(id) from group_sale_bonuses where product_id = 1) = 0) THEN
INSERT INTO group_sale_bonuses (member_id, product_id, quantity_counter, credit)
SELECT id, 1, 0, 0 FROM members
END IF
But this sql causes the errors.
I know there are solutions about Insert Ignore, Where Not Exists.
But these conditions checking are based on per each record. I have thousands of records in members table. I want to make condition checking just one time like in my above sql example.
By the way, I will use this Sql in Php web application.

Answer 1

You could just set the code in a WHERE clause instead of the IF.
INSERT INTO group_sale_bonuses(
member_id,
product_id,
quantity_counter,
credit)
SELECT
id, 1, 0, 0 FROM members
WHERE(
SELECT
count(id) FROM group_sale_bonuses
WHERE product_id = 1
) = 0;

Answer 2

This should do it for all product_id's
SELECT m.product_id, m.member_id FROM members AS m
LEFT JOIN group_sale_bonuses AS gsb ON gsb.product_id = m.product_id
WHERE gsb.product_id IS NULL ;
You can filter it to a specific product_id by adding to the where clause
SELECT m.product_id, m.member_id FROM members AS m
LEFT JOIN group_sale_bonuses AS gsb ON gsb.product_id = m.product_id
WHERE gsb.product_id IS NULL AND m.product_id = 1;
Take a look at this SQLfiddle: http://sqlfiddle.com/#!2/8482c/2