Famous SQL Interview Question | First Name , Middle Name and Last Name of a Customer

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Famous SQL Interview Question | First Name , Middle Name and Last Name of a Customer

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Күн бұрын

Description
In this SQL tutorial, we tackle a famous SQL interview question: how to extract the first name, middle name, and last name of a customer from a database. This is a common problem that tests your ability to handle string manipulation and understand the structure of names within a database. We'll walk you through the SQL query to efficiently retrieve these name components, ensuring you're well-prepared for your next technical interview.
Keywords
SQL interview question
SQL tutorial
SQL string manipulation
Extract names in SQL
SQL query example
SQL name extraction
SQL for beginners
SQL for interviews
SQL name components
Tags
#SQL
#SQLTutorial
#InterviewQuestions
#SQLInterview
#StringManipulation
#DatabaseQuery
#SQLExamples
#TechnicalInterview
#LearnSQL
#ProgrammingTutorials
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Пікірлер: 9

@sahilnaik3904 Ай бұрын

with col_transformed as ( select *, (len(name) - len(replace(name,' ',''))) as checker, CHARINDEX(' ',name) as first_space, CHARINDEX(' ',name,CHARINDEX(' ',name)+1) as second_space from test_names ) select name, case when checker = 0 then name else SUBSTRING(name,1,first_space-1) end as first_name, case when checker = 2 then SUBSTRING(name,first_space+1, (second_space - first_space)) else null end as middle_name, case when checker != 0 then case when checker = 1 then SUBSTRING(name,first_space+1, (len(name)-first_space)) else SUBSTRING(name,second_space+1, (len(name)-second_space)) end else null end as last_name from col_transformed;

@varunas9784 Ай бұрын

Thanks for sharing. Here's my approach on SQL server: =========================================== with cte2 as (select *, ROW_NUMBER() over(partition by name order by (select 1)) rn from names cross apply string_split(name,' ')) select Max(First_name) First_name, Max(Middle_name) Middle_name, Max(Last_name) Last_name from (select *, case when rn = 1 then value end as First_name, case when COUNT(name) over(partition by name) > 2 and rn = 2 then value end Middle_name, case when COUNT(name) over(partition by name) < 3 and rn = 2 then value when COUNT(name) over(partition by name) >= 3 and rn = 3 then value end Last_name from cte2) s1 group by name ===========================================

@rajanikanthgaja2916 Ай бұрын

Excellent

@sabesanj5509 Ай бұрын

Hi varun, Thanks for your query. Is your query gave the desired output??

@chandanpatra1053 Ай бұрын

I like your approach to make others understand after writing each query and that too step by step in excel...Great keep it up....looking forward to see 1 video in every two days.....Please bring quality level sql questions.👍👍

@rajanikanthgaja2916 Ай бұрын

Sure definitely thanks for your feedback plz subscribe and share my video's for better reach

@gajasrikanth4242 Ай бұрын

🎉🎉❤🎉🎉

@Tejasri15150 Ай бұрын

❤

@sabesanj5509 Ай бұрын

WITH cte AS ( SELECT customer_name, TRIM(value) AS part, ROW_NUMBER() OVER (PARTITION BY customer_name ORDER BY (SELECT 1)) AS rn FROM customers CROSS APPLY STRING_SPLIT(customer_name, ' ') ) SELECT customer_name, MAX(CASE WHEN rn = 1 THEN part END) AS first_name, MAX(CASE WHEN rn = 2 THEN part END) AS middle_name, MAX(CASE WHEN rn = 3 THEN part END) AS last_name FROM cte GROUP BY customer_name;