我收到错误:无法通过引用传递参数 2.....
I am getting the error: Cannot pass parameter 2 by reference in.....
在这一行...
$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 ??
它期望第二个参数是一个可以通过引用传递的变量.假设 $stmt1
是一个 PDO 语句,作为 docs 对于 bindparam 说
It expects the second paramter to be a variable which can be passed by reference. Assuming $stmt1
is a PDO statement then, as the docs for bindparam say
与 PDOStatement::bindValue() 不同,变量绑定为引用并且只会在 PDOStatement::execute() 被评估时叫.
Unlike PDOStatement::bindValue(), the variable is bound as a reference and will only be evaluated at the time that PDOStatement::execute() is called.
你的第二个参数是一个表达式 ($_SESSION['quantity'.$i] * $_SESSION['price'.$i]
) 不是一个变量.由于您现在似乎想要评估表达,我想您应该使用 bindValue()
代替.
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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