无法将内爆数组绑定到 mysql 准备好的语句中

您可以像这样生成类型字符串:

You can generate the types string like this:

$types = str_repeat('s', count($selected));

所有这些都是创建一个 s 的字符串，它的字符数与数组中的元素数一样多.

All this does is create a string of s's that is as many characters as the number of elements in the array.

然后你可以像这样使用 call_user_func_array() 绑定你的参数(注意我把括号放回 IN() 函数):

You would then bind your parameters using call_user_func_array() like this (notice I put the parenthesis back in for the IN() function):

if ($stmt = $mysqli->prepare("DELETE FROM email_addresses WHERE email_addresses IN (?)")) {
    call_user_func_array(array($stmt, "bind_param"), array_merge($types, $selected));

但是如果你尝试这个你会得到一个关于 mysqli_stmt::bind_param() 期望参数二通过引用传递的错误:

But if you try this you will get an error about mysqli_stmt::bind_param() expecting parameter two to be passed by reference:

警告:mysqli_stmt::bind_param() 的参数 2 应为引用，已给定值

Warning: Parameter 2 to mysqli_stmt::bind_param() expected to be a reference, value given

这有点烦人，但很容易解决.要解决此问题，您可以使用以下函数:

This is kind of annoying but easy enough to work around. To work around that you can use the following function:

function refValues($arr){ 
    $refs = array(); 
    foreach($arr as $key => $value) 
        $refs[$key] = &$arr[$key]; 
    return $refs; 
} 

它只是创建一个值数组，这些值是对 $selected 数组中值的引用.这足以让 mysqli_stmt::bind_param() 开心:

It just creates an array of values that are references to the values in the $selected array. This is enough to make mysqli_stmt::bind_param() happy:

if ($stmt = $mysqli->prepare("DELETE FROM email_addresses WHERE email_addresses IN (?)")) {
    call_user_func_array(array($stmt, "bind_param"), array_merge($types, refValues($selected)));

编辑

从 PHP 5.6 开始，您现在可以使用 ... 运算符来使这更简单:

As of PHP 5.6 you can now use the ... operator to make this even simpler:

$stmt->bind_param($types, ...$selected);

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