无法通过PHP PDO中的参考错误传递参数2 [英] Cannot pass parameter 2 by reference error in php PDO

问题描述

我遇到了错误: 无法通过引用将参数2传入.....

I am getting the error: Cannot pass parameter 2 by reference in.....

在此行...

$stmt1->bindParam(':value', $_SESSION['quantity'.$i] * $_SESSION['price'.$i], PDO::PARAM_STR);   

上面的代码有什么问题?

What is wrong with code above ??

推荐答案

它期望第二个参数是可以通过引用传递的变量.假设$stmt1是PDO语句，则如bindparam的 docs 所说的

It expects the second paramter to be a variable which can be passed by reference. Assuming $stmt1is a PDO statement then, as the docs for bindparam say

与PDOStatement :: bindValue()不同，该变量被绑定为引用 并且只会在PDOStatement :: execute()为 叫.

Unlike PDOStatement::bindValue(), the variable is bound as a reference and will only be evaluated at the time that PDOStatement::execute() is called.

您的第二个参数是表达式($_SESSION['quantity'.$i] * $_SESSION['price'.$i])而不是变量.既然您似乎现在想评估该表达式，我想您应该改用bindValue().

Your second param is an expression ($_SESSION['quantity'.$i] * $_SESSION['price'.$i]) not a variable. Since you appear to want to evaluate the exptression now, I guess you should used bindValue() instead.

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