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Databases

Posted by dtap
dtap
3,496 Points

SQL Finding Data that Matches Patterns - Challenge 1 of 2

SQL Basics Challenge Task 1 of 2. In the e-commerce database we have a products table. The columns are id, name, description and price. Find all the products where the pattern 't-shirt' can be found anywhere in the product name.

This was my query and it's saying it is incorrect. Anyone with insight so kind enough to help? Thanks folks

SELECT name AS "Product Name", description AS "Product Description" FROM products;

6 Answers

jcorum
jcorum
71,830 Points
jcorum
jcorum
71,830 Points

Thanks. Yes, they wanted something a bit different:

Task 1

SELECT * FROM products WHERE name LIKE '%t-shirt%'

Task 2

SELECT * FROM users WHERE first_name LIKE 'L%'
chase singhofen
chase singhofen
3,811 Points
chase singhofen
chase singhofen
3,811 Points

i kept putting the semi colon. very specific and detailed

jcorum
jcorum
71,830 Points
jcorum
jcorum
71,830 Points

Without a link to the challenge it's a bit of a guess, but try this:

SELECT name AS "Product Name", description AS "Product Description" FROM products WHERE name LIKE '%t-shirt&';

The % is a wild-card, and means any characters at all (even none). So this would get "Large t-shirt", "t-shirt", and "t-shirt with logo".

They may also want you to make it case-insensitive.

dtap
dtap
3,496 Points
dtap
dtap
3,496 Points

@jcorum - Thanks again!

Sandeep Krishnan
Sandeep Krishnan
9,730 Points
Sandeep Krishnan
Sandeep Krishnan
9,730 Points

Really didn't understand the difference between

SELECT * FROM user WHERE first_name LIKE "%L";

AND

SELECT * FROM user WHERE first_name LIKE "L%";

NiKole Maxwell
NiKole Maxwell
11,083 Points

%L ** - ending with the letter L. In a table of first_names it would return - Hal, Mel, and Saul **L% - beginning with the letter L - the same column would return - Linda, Larry, Lyric. %L% - would return names containing L, for example - Melissa, Alisha, Ally, Bartholomew

Examples of SQL wildcard use from W3schools :

**LIKE Operator Description**
WHERE CustomerName LIKE 'a%'    Finds any values that starts with "a"
WHERE CustomerName LIKE '%a'    Finds any values that ends with "a"
WHERE CustomerName LIKE '%or%'  Finds any values that have "or" in any position
WHERE CustomerName LIKE '_r%'   Finds any values that have "r" in the second position
WHERE CustomerName LIKE 'a_%_%' Finds any values that starts with "a" and are at least 3 characters in length
WHERE ContactName LIKE 'a%o'    Finds any values that starts with "a" and ends with "o"
dtap
dtap
3,496 Points
dtap
dtap
3,496 Points

@jcorum - Thanks for your help. I tried that query and I'm still getting the 'bummer' error.

The link to the challenge is here: https://teamtreehouse.com/library/sql-basics/finding-the-data-you-want/finding-data-that-matches-a-pattern-2

BT Murphy
BT Murphy
2,404 Points
BT Murphy
BT Murphy
2,404 Points

Thanks Nikole. Helped me on the next question after this one.