我正试图使用php的mysqli扩展为LIKE查询准备语句。但无论我怎么尝试，我总是得到这个错误:
Fatal error: Problem preparing query (SELECT f.*,r.slug FROM `foods` AS f INNER JOIN `resturants` AS r ON f.`rest_id` = r.`rest_id` WHERE f.`name` LIKE CONCAT('%',"f", '%')) 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 ''%',"f", '%')' at line 4 in /path/to/class.mysqli.php on line 462
我尝试了以下查询，但没有效果:
($s将是我要搜索的字符串。)
SELECT f.*,r.slug FROM `foods` AS f
INNER JOIN `resturants` AS r
ON f.`rest_id` = r.`rest_id`
WHERE f.`name` LIKE '%?%'
SELECT f.*,r.slug FROM `foods` AS f
INNER JOIN `resturants` AS r
ON f.`rest_id` = r.`rest_id`
WHERE f.`name` LIKE CONCAT('%', ?, '%')
SELECT f.*,r.slug FROM `foods` AS f
INNER JOIN `resturants` AS r
ON f.`rest_id` = r.`rest_id`
WHERE f.`name` LIKE CONCAT('%', {$s}, '%')
SELECT f.*,r.slug FROM `foods` AS f
INNER JOIN `resturants` AS r
ON f.`rest_id` = r.`rest_id`
WHERE f.`name` LIKE '%{$s}%'
:
sprintf("SELECT f.*,r.slug FROM `foods` AS f
INNER JOIN `resturants` AS r
ON f.`rest_id` = r.`rest_id`
WHERE f.`name` LIKE '%%%s%%'", $s)
请帮帮我，我快抓狂了
我会把LIKE后面的表达式移到变量:
$param = '%somestring%';
$sql = "SELECT f.*,r.slug FROM `foods` AS f
INNER JOIN `resturants` AS r
ON f.`rest_id` = r.`rest_id`
WHERE f.`name` LIKE ?"
更新:
也许这会有帮助
-- test.sql
CREATE TABLE supportContacts (
id int auto_increment primary key,
type varchar(20),
details varchar(30)
);
INSERT INTO supportContacts
(type, details)
VALUES
('Email', 'admin@sqlfiddle.com'),
('Twitter', '@sqlfiddle');
<?php
// test.php
$mysqli = new mysqli("localhost", "root", "root", "test");
$sql = 'SELECT type FROM supportContacts WHERE type LIKE ?'; // here is only ?, no %
$stmt = $mysqli->prepare($sql);
$type = 'E%'; // and here you can put % sign
$stmt->bind_param('s', $type);
$stmt->execute();
$stmt->bind_result($result);
$stmt->fetch();
var_dump($result);
您的%也需要用引号封装。