Win a copy of JDBC Workbook this week in the JDBC and Relational Databases forum
or A Day in Code in the A Day in Code forum!
  • Post Reply Bookmark Topic Watch Topic
  • New Topic
programming forums Java Mobile Certification Databases Caching Books Engineering Micro Controllers OS Languages Paradigms IDEs Build Tools Frameworks Application Servers Open Source This Site Careers Other all forums
this forum made possible by our volunteer staff, including ...
  • Campbell Ritchie
  • Paul Clapham
  • Jeanne Boyarsky
  • Junilu Lacar
  • Henry Wong
  • Ron McLeod
  • Devaka Cooray
  • Tim Cooke
Saloon Keepers:
  • Tim Moores
  • Stephan van Hulst
  • Frits Walraven
  • Tim Holloway
  • Carey Brown
  • Piet Souris
  • salvin francis
  • fred rosenberger

My exam cloud mock 3, LIKE

Ranch Foreman
Posts: 1897
  • Mark post as helpful
  • send pies
  • Quote
  • Report post to moderator

...The developer needs customers residing in Okwana Street only.
Which WHERE clause is correct?
A. WHERE cust.street LIKE 'Okwana_'
B. WHERE cust.street LIKE 'Okwana%'
C. WHERE cust.street LIKE 'Okwana-_-_'
D. WHERE cust.street LIKE 'Okwana-%-%'
E. WHERE cust.street LIKE 'Okwana-__-__'
F. WHERE cust.street LIKE 'Okwana-%%%-%%%%'
The correct answer is E.

The LIKE expression uses wildcard characters to search for strings that match the wildcard pattern. In this case, the query uses the LIKE expression and the % wildcard to find all phone numbers begin with the string ?Okwana-? WHERE cust.street LIKE 'Okwana-%%%-%%%%'.

I think the closest answer is B because we are searching for 'Okwana Street' without any '-' signs.
Creator of Enthuware JWS+ V6
Posts: 3341
Android Eclipse IDE Chrome
  • Mark post as helpful
  • send pies
  • Quote
  • Report post to moderator
This question and answer don't seem to match: very messy stuff...
So there I was, trapped in the jungle. And at the last minute, I was saved by this tiny ad:
Devious Experiments for a Truly Passive Greenhouse!
    Bookmark Topic Watch Topic
  • New Topic