URI Online Judge Solution 2625 CPF Validation Using PostgreSQL Query Language. Your company's communications managers want a report on the natural person customer data that is registered in the database. But the old report had a problem. customers CPF data came without validation.
So your job now is to select all the CPFs of all the customers, and apply a mask on the return of the data.
The CPF mask looks like: '000.000.000-00'.
Schema
customers| Column | Type |
| id (PK) | numeric |
| name | character varying (255) |
| street | character varying (255) |
| city | character varying (255) |
| state | char (2) |
| credit_limit | numeric |
natural_person| Column | Type |
| id_customers (FK) | numeric |
| cpf | char (14) |
Tables
customers| id | name | street | city | state | credit_limit |
| 1 | Nicolas Diogo Cardoso | Acesso Um | Porto Alegre | RS | 475 |
| 2 | Cecília Olivia Rodrigues | Rua Sizuka Usuy | Cianorte | PR | 3170 |
| 3 | Augusto Fernando Carlos Eduardo Cardoso | Rua Baldomiro Koerich | Palhoça | SC | 1067 |
| 4 | Nicolas Diogo Cardoso | Acesso Um | Porto Alegre | RS | 475 |
| 5 | Sabrina Heloisa Gabriela Barros | Rua Engenheiro Tito Marques Fernandes | Porto Alegre | RS | 4312 |
| 6 | Joaquim Diego Lorenzo Araújo | Rua Vitorino | Novo Hamburgo | RS | 2314 |
natural_person| id_customers | cpf |
| 1 | 26774287840 |
| 2 | 97918477200 |
Output Sample
| CPF |
| 267.742.878-40 |
| 979.184.772-00 |
URI 2625 Solution in SQL:
SELECT CONCAT(substr(cpf,1,3),'.',SUBSTR(cpf,4,3),'.',
SUBSTR(cpf,7,3),'-',SUBSTR(cpf,10,2)) "CPF"
FROM natural_person;
Comments
Post a Comment