如何将此 SQL 转换为准备好的语句？

Question

我需要帮助将此 SQL 转换为准备好的语句。这是我的搜索栏。我是初学者，希望能得到一些帮助。

这是我的SQL

$conn = mysqli_connect('localhost','root','','my_db');
            
$mysql = "SELECT * FROM catetable";
$bike_list = mysqli_query($conn,$mysql);

$catesql = "SELECT catename FROM catetable";
$cate_list = mysqli_query($conn,$catesql);

这就是我想更改为 Prepared Statement

if (isset($_GET['search']))
{
    
    $search = $_GET['search'];
    
    $searchlist = array();
    $lowersearchlist = array();
    $i = 0;
    while ($one_cate = mysqli_fetch_assoc($cate_list))
    {
        $searchlist[$i] = $one_cate['catename'];
        $lowersearchlist[$i] = strtolower($one_cate['catename']);
        $i++;
    }
    if (in_array($search,$searchlist) || in_array($search,$lowersearchlist))
    {
        header("Location:feature.php");
    }
    else
    {
        header("Location:index.php?error=true");
        
    }
}

Answer 1

编写一个与 WHERE 子句中的参数相匹配的查询。 MySQL 通常默认进行不区分大小写的比较，因此您无需获取所有行以不区分大小写地准确比较它们。

if (isset($_GET['search'])) {
    $stmt = $conn->prepare("SELECT COUNT(*) AS c FROM yourTable WHERE catename = ?");
    $stmt->bind_param("s", $_GET['search']);
    $stmt->execute();
    $result = $stmt->get_result();
    $row = $result->fetch_assoc();
    if ($row['c'] > 0) {
        header("Location: feature.php");
    } else {
        header("Location: index.php?error=true";
    }
}

如何将此 SQL 转换为准备好的语句？

How do I convert this SQL to prepared statement?

php

sql

mysqli

prepared-statement