让我的函数访问外部变量 [英] Giving my function access to outside variable

问题描述

我在外面有一个数组:

$myArr = array();

我想让我的函数访问它外部的数组，以便它可以向它添加值

I would like to give my function access to the array outside it so it can add values to it

function someFuntion(){
    $myVal = //some processing here to determine value of $myVal
    $myArr[] = $myVal;
}

如何为函数赋予变量正确的作用域?

How do I give the function the right scoping to the variable?

推荐答案

默认情况下，当您处于函数内部时，您无权访问外部变量.

By default, when you are inside a function, you do not have access to the outer variables.

如果您希望您的函数可以访问外部变量，则必须在函数内部将其声明为 global :

function someFuntion(){
    global $myArr;
    $myVal = //some processing here to determine value of $myVal
    $myArr[] = $myVal;
}

有关详细信息，请参阅变量范围.

For more informations, see Variable scope.

但请注意，使用全局变量不是一个好习惯:这样，您的函数就不再独立了.

But note that using global variables is not a good practice : with this, your function is not independant anymore.

一个更好的主意是让你的函数返回结果:

function someFuntion(){
    $myArr = array();       // At first, you have an empty array
    $myVal = //some processing here to determine value of $myVal
    $myArr[] = $myVal;      // Put that $myVal into the array
    return $myArr;
}

并像这样调用函数:

$result = someFunction();

您的函数还可以接受参数，甚至处理通过引用传递的参数:

function someFuntion(array & $myArr){
    $myVal = //some processing here to determine value of $myVal
    $myArr[] = $myVal;      // Put that $myVal into the array
}

然后，像这样调用函数:

Then, call the function like this :

$myArr = array( ... );
someFunction($myArr);  // The function will receive $myArr, and modify it

有了这个:

• 您的函数接收外部数组作为参数
• 并且可以修改它，因为它是通过引用传递的.
• 这比使用全局变量更好:您的函数是一个单元，独立于任何外部代码.

有关更多信息，您应该阅读函数 部分，尤其是以下子部分:

For more informations about that, you should read the Functions section of the PHP manual, and,, especially, the following sub-sections :

• 函数参数
• 返回值

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