Home > Database > Mysql Tutorial > How Can I Find Cars That Passed All Listed Tests?

How Can I Find Cars That Passed All Listed Tests?

Patricia Arquette
Release: 2025-01-14 10:40:43
Original
984 people have browsed it

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:

1

2

3

4

<code class="language-sql">SELECT carname

FROM PassedTest

GROUP BY carname

HAVING COUNT(DISTINCT testtype) = (SELECT COUNT(*) FROM PassedTest GROUP BY carname);</code>

Copy after login

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:

1

2

3

4

5

6

7

8

<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>

Copy after login

Alternative:

Another option is to use the JOIN operation:

1

2

3

4

5

<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>

Copy after login

The above is the detailed content of How Can I Find Cars That Passed All Listed Tests?. For more information, please follow other related articles on the PHP Chinese website!

source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Latest Articles by Author
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template