SQL Using Null

1. SELECT name FROM teacher WHERE dept IS NULL
2. SELECT teacher.name, dept.name FROM teacher INNER JOIN dept ON (teacher.dept=dept.id)
3. SELECT teacher.name, dept.name FROM teacher LEFT JOIN dept ON (teacher.dept=dept.id)
4. SELECT teacher.name, dept.name FROM teacher RIGHT JOIN dept ON (teacher.dept=dept.id)
5. SELECT name, COALESCE(mobile,'07986 444 2266') AS mobile FROM teacher
6. SELECT teacher.name, COALESCE(dept.name,'None') AS mob FROM teacher LEFT JOIN dept ON (teacher.dept=dept.id) 
7. SELECT COUNT(name) AS name,COUNT(mobile) AS mobile FROM teacher
8. SELECT dept.name, COUNT(teacher.name) FROM teacher RIGHT JOIN dept ON (teacher.dept=dept.id) GROUP BY dept.name
9. SELECT teacher.name, CASE WHEN teacher.dept=1 THEN 'Sci' WHEN teacher.dept=2 THEN 'Sci' ELSE 'Art' END dept FROM teacher
10. SELECT teacher.name, CASE WHEN teacher.dept=1 THEN 'Sci' WHEN teacher.dept=2 THEN 'Sci' WHEN teacher.dept=3 THEN 'Art' ELSE 'None' END dept FROM teacher

+ Numeric Examples
1. SELECT A_STRONGLY_AGREE FROM nss WHERE question='Q01' AND institution='Edinburgh Napier University' AND subject='(8) Computer Science'
2. SELECT institution, subject FROM nss WHERE question='Q15' AND score="100"
3. SELECT institution,score FROM nss WHERE question='Q15' AND score < 50 AND subject='(8) Computer Science'
4. SELECT subject,SUM(response) FROM nss WHERE question='Q22' AND subject IN ('(8) Computer Science','(H) Creative Arts and Design') GROUP BY subject
5. SELECT subject,SUM(response*A_STRONGLY_AGREE/100) FROM nss WHERE question='Q22' AND subject IN("(8) Computer Science","(H) Creative Arts and Design") GROUP BY subject
6. SELECT subject, ROUND(SUM(response*A_STRONGLY_AGREE)/SUM(response)) FROM nss WHERE question='Q22' AND subject IN('(8) Computer Science', '(H) Creative Arts and Design') GROUP BY subject
7. SELECT institution,ROUND(SUM(score*response)/SUM(response)) FROM nss WHERE question='Q22' AND (institution LIKE '%Manchester%') GROUP BY institution
8. NO CLUE