how to use LIKE clause in mysql when using with php

Question

i'm trying to use like in mysql from php and i write this code

$query = mysqli_query($connection, "SELECT * FROM items order by create_at desc where content LIKE '" . $content . "%'")
  or die(mysqli_error($connection));

but it says there is an error in my syntax it says like this

You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'where content LIKE 'c%'' at line 1

problem in ordering, SELECT * FROM TABLE WHERE CONDITION ORDER LIMIT. you should follow mysql order of statements. — Devsi Odedra
– Devsi Odedra, Commented Jan 9, 2019 at 7:02

Muhammad Tahir Qaiser · Accepted Answer · 2023-03-06 02:11:25Z

4

Order BY clause always come in the end, use WHERE clause first and then use ORDER BY

Solution would be this :

SELECT * FROM items where content LIKE 'yourVariable%' ORDER BY create_at DESC

edited Mar 6, 2023 at 2:11

answered Jan 9, 2019 at 7:00

Muhammad Tahir Qaiser

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2 Comments

Muhammad Tahir Qaiser Over a year ago

Logically speaking , we get data first and then sort it later!

Déjà vu Over a year ago

...except LIMIT

Naveed Ramzan · Accepted Answer · 2019-01-09 07:26:55Z

1

Your PHP code should be:

$query  = "SELECT * FROM items  where content LIKE '" . $content . "%' order by create_at desc";
$result = mysqli_query($connection, $query) or die(mysqli_error($connection));

Query clauses sequence should be :

Select
From
join
where
group by
order by
limit

answered Jan 9, 2019 at 7:26

Naveed Ramzan

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Comments

Peter Mortensen · Accepted Answer · 2023-06-28 11:22:15Z

0

$query = mysqli_query($connection, "SELECT * FROM items order by create_at desc where content LIKE '%.$content.%'") or die(mysqli_error($connection));

edited Jun 28, 2023 at 11:22

Peter Mortensen

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answered Jan 9, 2019 at 7:29

Erdz Wardex

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