MySQLi prepared statement complains that quot;only variables should be passed by referencequot;(MySQLi 准备好的语句抱怨“只应通过引用传递变量;)
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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