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171 lines
2.8 KiB
Markdown
171 lines
2.8 KiB
Markdown
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# Queries
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## Joins
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1. Get all batch names along with their instructor names.
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```sql
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SELECT
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b.name, i.first_name, i.last_name
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FROM
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batches b
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JOIN
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instructors i ON b.instructor_id = i.id;
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```
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2. Get all students along with their batch names if present else `NULL`.
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```sql
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SELECT
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s.first_name, s.last_name, b.name
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FROM
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students s
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LEFT JOIN
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batches b ON s.batch_id = b.id;
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```
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3. Get all students along with their batch names. Also, fetch all the batches which have no students.
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```sql
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SELECT
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s.first_name, s.last_name, b.name
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FROM
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students s
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RIGHT JOIN
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batches b ON s.batch_id = b.id
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```
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4. Get all the combinations of batches and instructors.
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```sql
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SELECT
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b.name, i.first_name, i.last_name
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FROM
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batches b, instructors i
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```
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5. Get all students with their instructors. If a student has no instructor, then show `NULL` for the instructor's name.
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```sql
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SELECT
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s.first_name, s.last_name, i.first_name, i.last_name
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FROM
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students s
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LEFT JOIN
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batches b ON s.batch_id = b.id
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LEFT JOIN
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instructors i ON b.instructor_id = i.id;
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```
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## Aggregation
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1. Get the maximum IQ in all students (Try without aggregation first).
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```sql
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SELECT
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first_name, iq
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FROM
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students
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ORDER BY iq DESC
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LIMIT 1;
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```
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2. Get the maximum IQ in all students (With aggregation).
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```sql
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SELECT
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MAX(IQ) AS 'IQ'
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FROM
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students;
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```
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3. Get the oldest batch from the batches table.
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```sql
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SELECT
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MIN(start_date) AS 'start date'
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FROM
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batches;
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```
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4. Fetch the number of batches that start with the word `Jedi`.
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```sql
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SELECT
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COUNT(id)
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FROM
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batches
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WHERE
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name LIKE 'Jedi%';
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```
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5. Get the average IQ of all students (Without using `AVG`)
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```sql
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SELECT
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SUM(iq) / COUNT(iq) as 'Average IQ'
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FROM
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students;
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```
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6. Get the average IQ of students in all batches.
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```sql
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SELECT
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AVG(IQ) AS 'IQ'
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FROM
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students
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WHERE
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batch_id IS NOT NULL;
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```
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7. Find the average IQ of students in each batch.
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```sql
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SELECT
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batch_id, AVG(iq)
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FROM
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students
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GROUP BY batch_id;
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```
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8. Find the total number of students in each batch.
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```sql
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SELECT
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b.name, COUNT(s.id)
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FROM
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batches b
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LEFT JOIN
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students s ON b.id = s.batch_id
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GROUP BY b.id;
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```
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9. Get the total number of batches taught by each instructor.
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```sql
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SELECT
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i.first_name, COUNT(b.id)
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FROM
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instructors i
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LEFT JOIN
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batches b ON i.id = b.instructor_id
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GROUP BY i.id;
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```
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10. Find the average IQ of students in batches with batch ID `1` and `2`.
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```sql
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SELECT
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batch_id, AVG(iq)
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FROM
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students
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WHERE batch_id IN (1, 2)
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GROUP BY batch_id;
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```
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11. Find count of students that are part of batches that have average IQ greater than `120`.
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```sql
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SELECT
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batch_id, AVG(iq) as avg_iq, COUNT(iq)
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FROM
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students
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GROUP BY batch_id
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HAVING avg_iq > 130;
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```
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