带有动态参数的动态选择 mysqli 查询返回错误与绑定变量的数量不匹配

Which I sincerely do not understand since it seems the count is alright. So this is what the code really looks like in its rude format. I can't see what I'm doing wrong.

//get variables
$mediaArray ='Facebook,Twitter,Twitch,';
$otherMedia = 'House';

//convert string to array
$socialArray = explode(',', $mediaArray)

//declare some variables to be used later
$andwhere = '';
$bp = '';
$socialmarray = ''

 //get every value from array of social media
foreach($socialArray as $socialmedia){

    $socialmarray .=$socialmedia.',';
    $andwhere .= " AND socialmedianame=?";
    $bp .='s';
}

//test strings
echo $wheres = $andwhere;//AND socialmedianame=? AND socialmedianame=? AND socialmedianame=?
echo $bip = $bp.'s';//ssss
echo $validarayy = rtrim($socialmarray,',');//Facebook,Twitter,Twitch

//select query
$selectquery = $conn->prepare("select * from mediaservices where socialmedianame=? $wheres");
$selectquery->bind_param("$bip",$otherMedia,$validarayy);
$selectquery->execute();
$resultquery = $selectquery->get_result();

推荐答案

因为:

1. 您正在使用用户提供的数据，您必须假设您的查询容易受到恶意注入攻击并且
2. 要构建到查询中的数据量是可变的/不确定的，并且
3. 您只是在单个表列上编写条件检查

您应该使用准备好的语句并将所有 WHERE 子句逻辑合并到一个 IN 语句中.

$mediaArray ='Facebook,Twitter,Twitch,';
$otherMedia = 'House';

$media = array_unique(explode(',', $mediaArray . $otherMedia));
$count = count($media);

$conn = new mysqli("localhost", "root", "", "myDB");
$sql = "SELECT * FROM mediaservices";
if ($count) {
    $stmt = $conn->prepare("$sql WHERE socialmedianame IN (" . implode(',', array_fill(0, $count, '?')) . ")");
    $stmt->bind_param(str_repeat('s', $count), ...$media);
    $stmt->execute();
    $result = $stmt->get_result();
} else {
    $result = $conn->query($sql);
}
foreach ($result as $row) {
    // access values like $row['socialmedianame']
}