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SQL Exercise: Number of substitutes for each stage of the tournament

SQL soccer Database: Basic Exercise-21 with Solution

21. From the following table, write a SQL query to count the number of substitutes during various stages of the tournament. Sort the result-set in ascending order by play-half, play-schedule and number of substitute happened. Return play-half, play-schedule, number of substitute happened.

Sample table: player_in_out


Sample Solution:

SELECT play_half,play_schedule,COUNT(*) 
FROM player_in_out 
WHERE in_out='I'
GROUP BY play_half,play_schedule
ORDER BY play_half,play_schedule,count(*) DESC;

Sample Output:

 play_half | play_schedule | count
-----------+---------------+-------
         1 | ET            |     4
         1 | NT            |     3
         2 | ET            |     5
         2 | NT            |   272
         2 | ST            |     9
(5 rows)

Relational Algebra Expression:

Relational Algebra Expression: Compute a list to show the number of substitute happened in various stage of play for the entire tournament.

Relational Algebra Tree:

Relational Algebra Tree: Compute a list to show the number of substitute happened in various stage of play for the entire tournament.

Practice Online



Sample Database: soccer

soccer database relationship structure

Query Visualization:

Duration:

Query visualization of Compute a list to show the number of substitute happened in various stage of play for the entire tournament - Duration

Rows:

Query visualization of Compute a list to show the number of substitute happened in various stage of play for the entire tournament - Rows

Cost:

Query visualization of Compute a list to show the number of substitute happened in various stage of play for the entire tournament - Cost

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Previous SQL Exercise: Number of players replaced in the extra time of play.
Next SQL Exercise: Number of shots taken in penalty shootout matches.

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SQL: Tips of the Day

Can a foreign key be NULL and/or duplicate?

First remember a Foreign key simply requires that the value in that field must exist first in a different table (the parent table). That is all an Foreign key is by definition. Null by definition is not a value. Null means that we do not yet know what the value is.

Ref: https://bit.ly/3uVu3OS