用于LIKE的MySqli prepare语句错误 [英] MySqli prepare statement error when used for LIKE

问题描述

我正在尝试使用php的mysqli扩展为LIKE查询准备一条语句.但是无论我尝试什么，我总是会收到此错误:

I'm trying to make a prepared statement for a LIKE query using php's mysqli extension. But no matter what I try, I always get this error:

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

我尝试了以下查询无济于事:
($s是我要搜索的字符串.)

I have tried the following queries to no avail:
(The $s would be my string to search against.)

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}%'

甚至:

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)

请帮助我，我感到沮丧.

Help me please, I'm getting frustrated.

推荐答案

我将在LIKE之后将表达式移动到变量:

I would move expression after LIKE to variable:

$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:

也许这会有所帮助

-- test.sql
CREATE TABLE supportContacts (
     id int auto_increment primary key, 
     type varchar(20), 
     details varchar(30)
);

INSERT INTO supportContacts
(type, details)
VALUES
('Email', '[e-mail]'),
('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);

这篇关于用于LIKE的MySqli prepare语句错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！