SQL error with LIKE - php

Question

SELECT * FROM `orders` WHERE id LIKE %1%
Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '%1%' at line 1
PHP
$sql = "SELECT * FROM `orders` ";
switch ($_POST['criteria']) {
case 'id':
$sql .= "WHERE id LIKE %" . (int) $_POST['search_input'] . "%";
break;
case 'OCR':
$sql .= "WHERE OCR LIKE %" . $db->quote($_POST['search_input']) . "%";
break;
case 'name':
$arr = explode(' ', $_POST['search_input']);
$firstname = $arr[0];
if (isset($arr[1])) {
$lastname = $arr[1];
} else {
$lastname = null;
}
$sql .= "WHERE firstname LIKE %" . $db->quote($firstname) . "% AND lastname LIKE %" . $db->quote($lastname) . "%";
break;
}
echo $sql;
$stmt = $db->query($sql);
$rows = $stmt->fetchAll();
The query is being outputted and it looks fine to me, but for some reason I am getting a syntax error ( I assume it is), however I can't seem to spot any problems?

Answer 1

LIKE operator is a string function. So you need to enclose it with single quotes(').
SELECT * FROM `orders` WHERE id LIKE '%1%';

Answer 2

You have quotes missing around your strings, so your quesries look something like:
SELECT * FROM orders where id LIKE %55%
instead of:
SELECT * FROM orders where id LIKE '%55%'
$sql = "SELECT * FROM `orders` ";
switch ($_POST['criteria']) {
case 'id':
$sql .= "WHERE id LIKE '%" . (int) $_POST['search_input'] . "%'";
break;
case 'OCR':
$sql .= "WHERE OCR LIKE '%" . $db->quote($_POST['search_input']) . "%'";
break;
case 'name':
$arr = explode(' ', $_POST['search_input']);
$firstname = $arr[0];
if (isset($arr[1])) {
$lastname = $arr[1];
} else {
$lastname = null;
}
$sql .= "WHERE firstname LIKE '%" . $db->quote($firstname) . "% AND lastname LIKE '%" . $db->quote($lastname) . "%'";
break;
}
echo $sql;
$stmt = $db->query($sql);
$rows = $stmt->fetchAll();
This answer should fix your problem but I strongly suggest you use = instead of LIKE since you are looking for unique orders identified by id.
Yhe way you script is currently written, if id is 55, you will get orders 55, 255, 5500, 1559...

Answer 3

Kindly write pattern in single qoute '' and like me sure
incorrect SELECT * FROM `orders` WHERE id LIKE %1%
correct- SELECT * FROM `orders` WHERE id LIKE '%1%'