Using character manipulation function display all the employees name whose name end with 'E'

Question

1.00/5 (1 vote)

See more:

PLZ FIND ME WAYS TO SOLVE ABOVE QUERRY

What I have tried:

SELECT ASCII('E')
FROM DUAL;

SELECT LAST_NAME
FROM EMPLOY_INFORMATION
WHERE SUBSTR(LAST_NAME, -1, 0) = CHR(69);

Posted 18-Sep-18 23:09pm

Member 13989907

Updated 19-Sep-18 1:45am

v2

Add a Solution

Comments

jaket-cp 19-Sep-18 6:04am

have you looked at LIKE?
Have a read of
https://www.techonthenet.com/oracle/like.php
Practice Exercise #1:

oh right - re-read your question - you need to use a string function of some sort

2 solutions

Add a Solution

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

OriginalGriff · Answer 1 · 2018-09-19T00:29:00

Solution 1

SQL

SELECT Last_Name 
FROM Employ_Information
WHERE RIGHT(Last_Name, 1) = 'E'

Or you could try:

SQL

SELECT Last_Name 
FROM Employ_Information
WHERE Last_Name LIKE '%E'

Posted 19-Sep-18 0:29am

OriginalGriff

Updated 19-Sep-18 0:31am

v2

kirthiga S · Answer 2 · 2018-09-19T01:45:00

Solution 2

select last_name from employ_information where ASCII(RIGHT(last_name,1))=69

Posted 19-Sep-18 1:45am

kirthiga S

Using character manipulation function display all the employees name whose name end with 'E'

2 solutions

Solution 1

Solution 2

Add your solution here

Preview 0

Using character manipulation function display all the employees name whose name end with 'E'

2 solutions

Solution 1

Solution 2

Add your solution here

Preview 0

Existing Members

...or Join us