Question

我有以下场景（简化）：

function changeFruit($fruit) {
    changeAgain($fruit);

}

function changeAgain($fruit) {
     $fruit = "Orange";
}

MAIN:

$fruit = "Apple";
changeFruit($fruit);
echo $fruit // Will show up as "Apple", How do I get it to show up as "Orange"??

编辑：忘了加入。情景不能使用退货声明 - 只是＆$变量

谢谢！马特穆勒

Answer 1

当您将不是对象的东西传递给PHP中的函数时，php会复制该函数以在函数中使用。

要使其不使用副本，您需要告诉PHP您正在传递引用。

这是通过＆amp;操作

function changeFruit(&$fruit) {
    changeAgain($fruit);

}

function changeAgain(&$fruit) {
     $fruit = "Orange";
}

$fruit = "Apple";
changeFruit($fruit);
echo $fruit;

使用函数的返回值会更加明智，更好的做法（因为这样可以更容易阅读）

function changeFruit($fruit) {
    return changeAgain($fruit);
}

function changeAgain($fruit) {
     // do something more interesting with$fruit here
     $fruit = "Orange";
     return $fruit;
}

$fruit = "Apple";
$fruit = changeFruit($fruit);
echo $fruit

Answer 2

function changeFruit($fruit) {
    return changeAgain($fruit);

}

function changeAgain($fruit) {
     return $fruit = "Orange";
}

MAIN:

$fruit = "Apple";
$fruit = changeFruit($fruit);
echo $fruit;

希望有所帮助！

注意：从changeAgain函数返回并覆盖$ fruit = changeFruit（$ fruit）;

Answer 3

您没有从函数中返回值。试试这个：

function changeFruit($fruit) {
    return changeAgain($fruit);

}

function changeAgain($fruit) {
     $fruit = "Orange";
     return $fruit;
}

MAIN:

$fruit = "Apple";
$fruit = changeFruit($fruit);

PHP功能混乱

3 个答案: