Computer Science
Table EXAM_RESULTS
STU ID | FNAME | LNAME | EXAM ID | EXAM_SCORE |
---|---|---|---|---|
10 | LAURA | LYNCH | 1 | 90 |
10 | LAURA | LYNCH | 2 | 85 |
11 | GRACE | BROWN | 1 | 78 |
11 | GRACE | BROWN | 2 | 72 |
12 | JAY | JACKSON | 1 | 95 |
12 | JAY | JACKSON | 2 | 92 |
13 | WILLIAM | BISHOP | 1 | 70 |
13 | WILLIAM | BISHOP | 2 | 100 |
14 | CHARLES | PRADA | 2 | 85 |
What SQL statement do we use to print out the record of all students whose last name starts with 'L'?
- SELECT * FROM EXAM_RESULTS WHERE LNAME LIKE 'L%' ;
- SELECT * FROM EXAM_RESULTS WHERE LNAME LIKE 'L';
- SELECT * FROM EXAM_RESULTS WHERE LNAME 'L';
- SELECT * FROM EXAM_RESULTS WHERE LNAME <> 'L';
Answer
SELECT * FROM EXAM_RESULTS WHERE LNAME LIKE 'L%' ;
Output
+--------+-------+-------+---------+------------+
| stu_id | fname | lname | exam_id | exam_score |
+--------+-------+-------+---------+------------+
| 10 | LAURA | LYNCH | 1 | 90 |
| 10 | LAURA | LYNCH | 2 | 85 |
+--------+-------+-------+---------+------------+
Explanation
SELECT * FROM EXAM_RESULTS WHERE LNAME LIKE 'L%';
— The LIKE operator is used for pattern matching in SQL. '%' is a wildcard character that matches zero or more characters. 'L%' specifies that the last name (LNAME) should start with 'L' followed by zero or more characters. TheSELECT *
statement retrieves all columns from the EXAM_RESULTS table for the matching records.SELECT * FROM EXAM_RESULTS WHERE LNAME LIKE 'L';
— This query attempts to select all columns (*) from the EXAM_RESULTS table where the last name (LNAME) is exactly equal to 'L'. However, when using the LIKE operator in SQL for pattern matching, we use wildcard characters (%) to represent unknown parts of a string.SELECT * FROM EXAM_RESULTS WHERE LNAME 'L';
— This statement contains a syntax error. In SQL, when using the WHERE clause to filter records based on a specific condition, we need to use comparison operators or functions to define the condition properly.SELECT * FROM EXAM_RESULTS WHERE LNAME <> 'L';
— This query retrieves records where the last name is not equal to 'L'. It does not specifically look for last names starting with 'L', so it's not the correct option for the given requirement.
Related Questions
Table EXAM_RESULTS
STU ID FNAME LNAME EXAM ID EXAM_SCORE 10 LAURA LYNCH 1 90 10 LAURA LYNCH 2 85 11 GRACE BROWN 1 78 11 GRACE BROWN 2 72 12 JAY JACKSON 1 95 12 JAY JACKSON 2 92 13 WILLIAM BISHOP 1 70 13 WILLIAM BISHOP 2 100 14 CHARLES PRADA 2 85 What SQL statement do we use to find the average exam score for EXAM_ID = 1?
- SELECT AVG(EXAM_SCORE) FROM EXAM_RESULTS;
- SELECT AVG(EXAM_SCORE) FROM EXAM_RESULTS GROUP BY EXAM_ID WHERE EXAM_ID = 1;
- SELECT AVG(EXAM_SCORE) FROM EXAM_RESULTS GROUP BY EXAM_ID HAVING EXAM_ID = 1;
- SELECT COUNT(EXAM_SCORE) FROM EXAM_RESULTS WHERE EXAM_ID = 1;
Table EXAM_RESULTS
STU ID FNAME LNAME EXAM ID EXAM_SCORE 10 LAURA LYNCH 1 90 10 LAURA LYNCH 2 85 11 GRACE BROWN 1 78 11 GRACE BROWN 2 72 12 JAY JACKSON 1 95 12 JAY JACKSON 2 92 13 WILLIAM BISHOP 1 70 13 WILLIAM BISHOP 2 100 14 CHARLES PRADA 2 85 Which SQL statement do we use to find out how many students took each exam?
- SELECT COUNT(DISTINCT STU_ID) FROM EXAM_RESULTS GROUP BY EXAM_ID;
- SELECT EXAM_ID, MAX(STU_ID) FROM EXAM_RESULTS GROUP BY EXAM_ID;
- SELECT EXAM_ID, COUNT(DISTINCT STU_ID) FROM EXAM_RESULTS GROUP BY EXAM_ID;
- SELECT EXAM_ID, MIN(STU_ID) FROM EXAM_RESULTS GROUP BY EXAM_ID;
Table EXAM_RESULTS
STU ID FNAME LNAME EXAM ID EXAM_SCORE 10 LAURA LYNCH 1 90 10 LAURA LYNCH 2 85 11 GRACE BROWN 1 78 11 GRACE BROWN 2 72 12 JAY JACKSON 1 95 12 JAY JACKSON 2 92 13 WILLIAM BISHOP 1 70 13 WILLIAM BISHOP 2 100 14 CHARLES PRADA 2 85 What is the result of the following SQL statement ?
SELECT MAX(EXAM_SCORE) FROM EXAM_RESULTS GROUP BY EXAM_ID HAVING EXAM_ID = 1;
- 90
- 85
- 100
- 95
Given the following table :
Table : CLUB
COACH-ID COACHNAME AGE SPORTS DATOFAPP PAY SEX 1 KUKREJA 35 KARATE 27/03/1996 1000 M 2 RAVINA 34 KARATE 20/01/1998 1200 F 3 KARAN 34 SQUASH 19/02/1998 2000 M 4 TARUN 33 BASKETBALL 01/01/1998 1500 M 5 ZUBIN 36 SWIMMING 12/01/1998 750 M 6 KETAKI 36 SWIMMING 24/02/1998 800 F 7 ANKITA 39 SQUASH 20/02/1998 2200 F 8 ZAREEN 37 KARATE 22/02/1998 1100 F 9 KUSH 41 SWIMMING 13/01/1998 900 M 10 SHAILYA 37 BASKETBALL 19/02/1998 1700 M Give the output of following SQL statements :
- SELECT COUNT(DISTINCT SPORTS) FROM Club ;
- SELECT MIN(Age) FROM CLUB WHERE Sex = 'F' ;
- SELECT AVG(Pay) FROM CLUB WHERE Sports = 'KARATE' ;
- SELECT SUM(Pay) FROM CLUB WHERE Datofapp > '1998-01-31' ;