php - LIKE 的 SQL 错误

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();

正在输出查询，对我来说它看起来很好，但由于某种原因我收到语法错误(我假设是)，但我似乎无法发现任何问题？

Answer 1

LIKE运算符是一个字符串函数。所以你需要用单引号(')把它括起来。

SELECT * FROM `orders` WHERE id LIKE '%1%';