MySQLi 准备好的语句抱怨“只应通过引用传递变量"； [英] MySQLi prepared statement complains that "only variables should be passed by reference"

问题描述

代码:

$stmt->bind_param("s", md5($input['user'] . $config['salt']));

PHP 错误信息:

只有变量应该通过引用传递

Only variables should be passed by reference

我一直在研究这个项目，但我现在被困住了.我是 PHP 新手.怎么办?

I've been working on this project but I am stuck now. I am new to PHP. What to do?

推荐答案

感谢您使用 MySQLi 准备好的语句！他们很痛苦，但值得.

Thanks for using MySQLi prepared statements! They're a pain, but it's worth it.

bind_param 通过引用.它通过查看您传递的变量并直接指向内脏来实现这一点.

bind_param takes values by reference. It does this by looking at the variable you're passing and pointing at the innards directly.

在您的调用中，您将返回函数调用的字符串结果 - 在本例中为 md5.因为没有涉及变量，所以没有内脏可指点.PHP 抱怨无法通过引用传递数据.

In your call, you're returning the string result of a function call - md5 in this case. Because there's no variable involved, there are no innards to point to. PHP is whining about not being able to pass the data by reference as a result.

您需要将函数调用的结果粘贴到一个变量中，然后将该变量传递给绑定.

You will need to stick the result of the function call into a variable, then pass that variable into the bind instead.

大警告！ md5 不再是一个安全的散列，并且应该不用于存储密码.如果有机会，您应该更新为更好的哈希格式，例如 bcrypt、PBKDF2、scrypt 等.

BIG FAT WARNING! md5 is not a secure hash any longer, and should not be used to store passwords. When you get the chance, you should update to a better hash format, such as bcrypt, PBKDF2, scrypt, etc.

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