PDO 通过引用通知?-php问题

PDO pass by reference notice?(PDO 通过引用通知?)

本文介绍了PDO 通过引用通知?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

这个:

$stmt = $dbh->prepare("SELECT thing FROM table WHERE color = :color");
$stmt->bindParam(':color', $someClass->getColor());
$stmt->execute();

产生这个:

运行时通知
只应传递变量参考

Runtime notice
Only variables should be passed by reference

虽然它仍然执行.

这个:

$stmt = $dbh->prepare("SELECT thing FROM table WHERE color = :color");
$tempColor = $someClass->getColor();
$stmt->bindParam(':color',$tempColor);
$stmt->execute();

毫无怨言地运行.

我不明白其中的区别?

推荐答案

bindParam 的第二个参数是一个变量参考.由于函数返回不能被引用，它不能严格满足bindParam参数的需求(PHP会配合你，这里只会发出警告).

The second parameter of bindParam is a variable reference. Since a function return cannot be referenced, it fails to strictly meet the needs of the bindParam parameter (PHP will work with you though and will only issue a warning here).

为了更好地理解，这里有一个例子:这段代码将产生与你的第二个例子相同的结果:

To get a better idea, here's and example: this code will produce the same results as your second example:

$stmt = $dbh->prepare("SELECT thing FROM table WHERE color = :color");
$tempColor = NULL; // assigned here
$stmt->bindParam(':color',$tempColor);
$tempColor = $someClass->getColor(); // but reassigned here
$stmt->execute();

这在函数返回的情况下是不可能的.

That won't be possible with a function return.

这篇关于PDO 通过引用通知?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！