How Can I Find Cars That Passed All Listed Tests?

Find cars that passed all tests from the test list

This task requires identifying cars from the "cars" table that have successfully passed all tests listed in the "passedtest" table. While a simple IN statement can retrieve cars that pass any single test, the requirement is to ensure that each car passes all tests in the list.

To achieve this, we can use a combination of aggregation (GROUP BY, HAVING) and set theory.

Solution:

<code class="language-sql">SELECT carname
FROM PassedTest
GROUP BY carname
HAVING COUNT(DISTINCT testtype) = (SELECT COUNT(*) FROM PassedTest GROUP BY carname);</code>

Explanation:

• Subquery (SELECT COUNT(*) FROM PassedTest GROUP BY carname) returns the total number of unique tests per car.
• HAVING COUNT(DISTINCT testtype) = (SELECT COUNT(*) FROM PassedTest GROUP BY carname) condition checks whether the count of different test types of a car is equal to the total number of tests, ensuring that the car has passed all tests.

To include data from the "cars" table, you can use the inner statement:

<code class="language-sql">SELECT *
FROM cars
WHERE carname IN (
    SELECT carname
    FROM PassedTest
    GROUP BY carname
    HAVING COUNT(DISTINCT testtype) = (SELECT COUNT(*) FROM PassedTest GROUP BY carname)
);</code>

Alternative:

Another option is to use the JOIN operation:

<code class="language-sql">SELECT DISTINCT c.carname
FROM cars c
JOIN PassedTest pt ON c.carname = pt.carname
GROUP BY c.carname
HAVING COUNT(*) = (SELECT COUNT(*) FROM PassedTest GROUP BY carname);</code>

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