Difference between revisions of "Using Null Quiz"

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Using Null Quiz
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Test your understanding of the NULL value
 
+
 
<div class='ref_section' style='background:none'>
 
<div class='ref_section' style='background:none'>
 
<table class='zoo db_ref'>
 
<table class='zoo db_ref'>
Line 9: Line 8:
 
<th>name</th>  
 
<th>name</th>  
 
<th>phone</th>
 
<th>phone</th>
<th>mobile</th>
 
 
</tr>
 
</tr>
 
<tr>
 
<tr>
Line 16: Line 14:
 
<td>Shrivell</td>
 
<td>Shrivell</td>
 
<td align='right'>2753</td>
 
<td align='right'>2753</td>
<td>07986 555 1234</td>
 
 
</tr>
 
</tr>
 
<tr>
 
<tr>
Line 23: Line 20:
 
<td>Throd</td>
 
<td>Throd</td>
 
<td align='right'>2754</td>
 
<td align='right'>2754</td>
<td>07122 555 1920</td>
 
 
</tr>
 
</tr>
 
<tr><td align='right'>103</td>
 
<tr><td align='right'>103</td>
 
<td align='right'>1</td>
 
<td align='right'>1</td>
 
<td>Splint</td>
 
<td>Splint</td>
<td align='right'>2293</td>
+
<td align='right'></td>
<td></td>
+
 
</tr>
 
</tr>
 
<tr><td align='right'>104</td>
 
<tr><td align='right'>104</td>
 
<td align='right'></td>
 
<td align='right'></td>
 
<td>Spiregrain</td>
 
<td>Spiregrain</td>
<td align='right'>3287</td>
+
<td align='right'></td>
<td></td>
+
 
</tr>
 
</tr>
 
<tr>
 
<tr>
 
<td align='right'>105</td>
 
<td align='right'>105</td>
 
<td align='right'>2</td>
 
<td align='right'>2</td>
<td>Cutflower</td>
+
<td>Cutflower </td>
 
<td align='right'>3212</td>
 
<td align='right'>3212</td>
<td>07996 555 6574</td>
 
 
</tr>
 
</tr>
 
<tr>
 
<tr>
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<td align='right'></td>
 
<td align='right'></td>
 
<td>Deadyawn</td>
 
<td>Deadyawn</td>
<td align='right'>3345</td>
+
<td align='right'></td>
<td></td>
+
</tr>
+
<tr>
+
<td colspan='5'>...</td>
+
 
</tr>
 
</tr>
 +
<tr><td colspan='4'></td></tr>
 
</table>
 
</table>
 
<table class='zoo db_ref'>
 
<table class='zoo db_ref'>
Line 73: Line 63:
 
<td>Engineering</td>
 
<td>Engineering</td>
 
</tr>
 
</tr>
<tr>
+
<tr><td colspan='2'></td></tr>
<td colspan='2'>...</td>
+
</tr>
+
 
</table>
 
</table>
 
</div>
 
</div>
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- set dept value of all teachers without department to 0
 
- set dept value of all teachers without department to 0
  
{Query: <code>SELECT name, CASE WHEN phone = 2752 THEN 'two' WHEN phone = 2753 THEN 'three' WHEN phone = 2754 THEN 'four' END AS digit FROM teacher</code> shows following 'digit':
+
{Query:  
 +
<syntaxhighlight lang='sql'>
 +
SELECT name,
 +
      CASE WHEN phone = 2752 THEN 'two'
 +
            WHEN phone = 2753 THEN 'three'
 +
            WHEN phone = 2754 THEN 'four'
 +
            END AS digit
 +
  FROM teacher</syntaxhighlight>
 +
 +
shows following 'digit':
 
|type="()"}
 
|type="()"}
 
+ 'four' for Throd
 
+ 'four' for Throd

Latest revision as of 16:08, 29 August 2013

Test your understanding of the NULL value

teacher
id dept name phone
101 1 Shrivell 2753
102 1 Throd 2754
103 1 Splint
104 Spiregrain
105 2 Cutflower 3212
106 Deadyawn
dept
id name
1 Computing
2 Design
3 Engineering

1. Select the code which uses a JOIN correctly.

 SELECT teacher.name, dept.name FROM teacher JOIN dept ON (dept = id)
 SELECT teacher.name, dept.name FROM teacher, dept INNER JOIN ON (teacher.dept = dept.id)
 SELECT teacher.name, dept.name FROM teacher, dept JOIN WHERE(teacher.dept = dept.id)
 SELECT teacher.name, dept.name FROM teacher OUTER JOIN dept ON dept.id
 SELECT teacher.name, dept.name FROM teacher LEFT OUTER JOIN dept ON (teacher.dept > dept.id)

2. Select the correct statement that shows the name of department which employs Cutflower

 SELECT dept.name FROM teacher JOIN dept ON (dept.id = (SELECT dept FROM teacher WHERE name = 'Cutflower'))
 SELECT dept.name FROM teacher JOIN dept ON (dept.id = teacher.dept) WHERE dept.id = (SELECT dept FROM teacher HAVING name = 'Cutflower')
 SELECT dept.name FROM teacher JOIN dept ON (dept.id = teacher.dept) WHERE teacher.name = 'Cutflower'
 SELECT dept.name FROM teacher JOIN dept WHERE dept.id = (SELECT dept FROM teacher WHERE name = 'Cutflower')
 SELECT name FROM teacher JOIN dept ON (id = dept) WHERE id = (SELECT dept FROM teacher WHERE name = 'Cutflower')

3. Select out of following the code which uses a JOIN to show a list of all the departments and number of employed teachers

 SELECT dept.name, COUNT(*) FROM teacher LEFT JOIN dept ON dept.id = teacher.dept
 SELECT dept.name, COUNT(teacher.name) FROM teacher, dept JOIN ON dept.id = teacher.dept GROUP BY dept.name
 SELECT dept.name, COUNT(teacher.name) FROM teacher JOIN dept ON dept.id = teacher.dept GROUP BY dept.name
 SELECT dept.name, COUNT(teacher.name) FROM teacher LEFT OUTER JOIN dept ON dept.id = teacher.dept GROUP BY dept.name
 SELECT dept.name, COUNT(teacher.name) FROM teacher RIGHT JOIN dept ON dept.id = teacher.dept GROUP BY dept.name

4. Using SELECT name, dept, COALESCE(dept, 0) AS result FROM teacher on teacher table will:

display 0 in result column for all teachers
display 0 in result column for all teachers without department
do nothing - the statement is incorrect
set dept value of all teachers to 0
set dept value of all teachers without department to 0

5. Query:

SELECT name,
       CASE WHEN phone = 2752 THEN 'two'
            WHEN phone = 2753 THEN 'three'
            WHEN phone = 2754 THEN 'four'
            END AS digit
  FROM teacher

shows following 'digit':

'four' for Throd
NULL for all teachers
NULL for Shrivell
'two' for Cutflower
'two' for Deadyawn

6. Select the result that would be obtained from the following code:

 SELECT name, 
      CASE 
       WHEN dept 
        IN (1) 
        THEN 'Computing' 
       ELSE 'Other' 
      END 
  FROM teacher
Table-A
ShrivellComputing
ThrodComputing
SplintComputing
SpiregrainOther
CutflowerOther
DeadyawnOther
Table-B
ShrivellComputing
ThrodComputing
SplintComputing
SpiregrainComputing
CutflowerComputing
DeadyawnComputing
Table-C
ShrivellComputing
ThrodComputing
SplintComputing
Table-D
Spiregrain Other
Cutflower Other
Deadyawn Other
Table-E
Shrivell 1
Throd 1
Splint 1
Spiregrain 0
Cutflower 0
Deadyawn 0
Table-A
Table-B
Table-C
Table-D
Table-E

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