there is a table EMPLOYEE contains 3 columns (name, email, job) in sql:
+--------------------------+--------------------------------------------+-----------+
| name | email | job |
+--------------------------+--------------------------------------------+-----------+
| Jack Smith | [Jack11@abcd.com] | Teacher
|
| Rock Jol | [Rock.jol@jfhro.com] |
cashier |
| Andrew stanley | [ANDREW80@hfwjs.com] | student
|
+--------------------------+--------------------------------------------+-----------+
create second table EMPLOYEE2 which has three columns as follows (in sql):
Jack Smith Jack11@abcd.com Teacher
Hint: can use update EMPLOYEE2 set email = ....
Solution:
Creating employee table and inserting rows:
create table employee(name char(100),email varchar(100),job char(20));
insert into employee values ("Jack Smith","[Jack11@abcd.com]","Teacher"),("Rock Jol","[Rock.jol@fhro.com]","cashier"),("Andrew Stanley","[ANDREW80@hfwjs.com]","student");
Creating employee2 table:
create table employee2(name char(100),email varchar(100),job char(20));
Here insert rows into employee2 table from employee table by removing bracets [] in email column.By using substr() function we can achieve that task.
Inserting rows to employee2 table:
insert into employee2 (name,email,job) select name,substr(email,2,length(email)-2),job from employee;
Here substr() function takes the email string from 2nd index and then it will extract length(email)-2 characters it will give last before position of the string.
Code and Output Screenshots:

Note: if you have any queries please post a comment thanks a lot..always available to help you..
there is a table EMPLOYEE contains 3 columns (name, email, job) in sql: +--------------------------+--------------------------------------------+-----------+ | name...
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DROP TABLE EMPLOYEE;
DROP TABLE JOB;
DROP TABLE EMP;
DROP TABLE EMP_1;
DROP TABLE EMP_2;
CREATE TABLE JOB(JOB_CODE CHAR (3) PRIMARY KEY, JOB_DESCRIPTION
VARCHAR (20) NOT NULL,JOB_CHG_HOUR NUMBER (5,2) NOT
NULL,JOB_LAST_UPDATE DATE NOT NULL);
INSERT INTO JOB
VALUES('500','Programmer','35.75','20-Nov-2017');
INSERT INTO JOB VALUES('501','System
Analyst','96.75','20-Nov-2017');
INSERT INTO JOB VALUES('502','Database
Designer','125.00','24-Mar-2018');
CREATE TABLE EMPLOYEE(EMP_NUM CHAR (3) PRIMARY KEY,EMP_LNAME
VARCHAR (15) NOT NULL,EMP_FNAME VARCHAR (15) NOT NULL, EMP_INITIAL
CHAR (1),EMP_HIREDATE DATE NOT NULL,JOB_CODE CHAR (3), EMP_YEARS
NUMBER (2),FOREIGN KEY (JOB_CODE) REFERENCES JOB (JOB_CODE));
INSERT...
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