pt-23109958page 3 essn: char(9) sex: char(1) o relationship varcharf8) fname: va

Question

pt-23109958page 3 essn: char(9) sex: char(1) o relationship varcharf8) fname: varchar(15 o minit : varchar(1) oIname:varchar(15) assn char(9) a bdate: date a address: varchar(50) o sex: char(1) a salary decimal(102)dnumber o superssn: char(9) #dno int(4) essn char(9) a pno : int(4) # hours, decinlaK41) e pname varchar(25) a prumber int(4) e plocation varchar(15) ndnum int(4) o dname:varchar(25) int(4) mgrssn:char(9) o mgrstartdate: date a dnumber: int(4) dlocation varchar(15) Based this schema, answer the following questions 1. Here are the results for two queries against the database above ariaDB [company]> select count() from employee; count (I 43 1 row in set (0.00 sec) Ma riaDB [company]> select count() from employee where superssn is not null; count(I 34 1 row in set (.00 sec) 1) select z.ssn, y superssn from employee z employee y has 2) select z.ssn, y superssn from employee z NATURAL JOIN employee y has 3) select z.ssn, y superssn from employee z cROSS JOIN employee y, has 4) select z.ssn, y superssn from employee z LEFT JOIN employee y ON z ssney ssn, has 5) select z.ssn, y superssn from employee z RIGHT OUTER JOIN employee y ON z ssnay ssn, has 6) (select z.ssn, y superssn from employee z LEFT JOIN employee y ON z ssnay ssn) union (select z ssn, y superssn from employee z RIGHT OUTER JOIN employee y ON z ssney ssn) records records records records records

Explanation / Answer

1. This Query has 1849 records.

There are total 43 records in the Employee table. so, It generates 43*43 records

2. This Query has 34 records.

It removes rows with null columns.

3. This Query has 1849 records.

It is same as 1st query.

4. This Query has 43 records.

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