Search string in Oracle SQL

Question

The question is about searching name starting with letter 'G' in employees table. I am using Oracle 12c Version.

I came across the answer:

select * 
from employees 
where first_name >= 'G' 
  and first_name < 'H';

Could you please help me to understand the logic behind this.

they're just string comparisons, which means 'g' < 'harrison' is TRUE, because g comes before h in the alphabet. the length of the strings is irrelevant. the characters are compared in lock step, as soon as there's s mismatch, you get the comparison results. — Marc B
– Marc B, Commented Sep 9, 2016 at 20:19

Rahul · Accepted Answer · 2016-09-09 20:23:36Z

2

Other than what is being commented, you can as well use LIKE operator since you are eventually trying to get all first_name starting with G

select * from employees where first_name like 'G%'

answered Sep 9, 2016 at 20:23

Rahul

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1 Comment

Thanks Marc & Rahul

XING · Accepted Answer · 2016-09-12 09:17:15Z

1

You can use regexp as well to search your names starting with G as below:

select * from employees 
WHERE REGEXP_LIKE (first_name, '^G','i');

here parameter 'i' will ignore the case of the name.

answered Sep 12, 2016 at 9:09

XING

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