将全局变量和参数传递给 PHP 中的函数

Question

您好，感谢您的光临，

我想将变量 ($user) 从前一个函数传递到另一个函数，但我需要使用新函数的参数来传递将要执行的值渲染这个新的。

有什么方法可以将变量从另一个函数传递到一个只需要三个参数的新函数，并且它们都不是前一个函数中的变量？

示例：

function my_function($country, $age, $colour) {
  if ($user = true) {
    echo "User is from " . $country . " and " . $age . " and his favourite colour is " . $colour; 
  }
}

my_function("italy", 19, "red");

如果我将 my_function: 放入函数内部，它会起作用，

global $user;

但我相信使用全局变量不是一个好的做法。

关于如何将其作为参数传递的任何想法？我是否应该将其添加为新函数参数中 $colour 之后的另一个变量？

非常感谢您的帮助:)

Answer 1

您可以将其作为参数传递，或者更好地这样做：

if ($user) my_function("italy", 19, "red");

因为您不必在该函数内使用 $user 变量。

Answer 2

您可以使用此函数，但最佳实践是使用类。
即如果你调用 my_function("italy", 19, "red"), $user 默认为 false

function my_function($country, $age, $colour, $user=false) {
if ($user == true) {
echo "User is from " $country . "and " . $age . " and his favourite colour is " . $colour; 
}
}

my_function("italy", 19, "red",true);

Answer 3

好吧，全局变量是一种不好的做法，但在您的情况下是可行的。

我绝对会建议您研究/使用面向对象的方法。

确实有很多不同的方法可以实现您的尝试。

一种方法是将代码封装到对象中。

<?php

class My_Cool_Class {
    public $user = false;

    public function myFunction($country, $age, $color) {
        if ($this->user)
        echo "User is from {$country} and {$age} years old and his favourite colour is {$color}"; 
    }
}

$class = new My_Cool_Class();
$class->user = new User();
$class->myFunction("Italy", 19, "red");

或者您可以实现单例模式，以便从任何地方轻松访问您的对象/函数。

<?php

class My_Cool_Class {
    public $user = false;

    protected static $_instance;

    public function getIntance() {
        if(!self::$_instance)
            self::$_instance = new self();

        return self::$_instance;
    }

    public function setUser($user) {
        $this->user = $user;
    }

    public function myFunction($country, $age, $color) {
        if ($this->user)
            echo "User is from {$country} and {$age} years old and his favourite colour is {$color}"; 
    }
}


//Set User from anywhere with this
My_Cool_Class::getInstance()->setUser($user);


//Call your function anywhere with this.
My_Cool_Class::getInstance()->myFunction("Italy", 19, "red");

Answer 4

如果你以前的函数是这样的：

/**
 * Callback function for so_user_data() that tells if we want to give you info about the user or not.
 * @param (string) $user | Accepts a Username as input
 * @return (boolean) | true if the User is 'Rob'
 */
function so_user_fn( $user )
{
    $allowed_users = array(
         'Rob'
        ,'Jay'
        ,'Silent Bob'
    );
    if ( in_array $user, $allowed_users ) )
        return true;
    // false if not 'Rob'
    return false;
}

/**
 * Shows the country, age & fav. colour of a user or denies displaying this information
 * if the user is not a public v.i.p. or someone other we want to give you info about.
 * @uses so_user_fn() | used to determine if the user is 'Rob'
 * @param (string) $user | Username
 * @param (string) $country | (optional) Country of origin
 * @param (integer) $age | (optional) Age of the user
 * @param (string) $colour | (optional) Fav. Colour
 */
function so_user_data( $user, $country = 'an unknown country', $age = 'unknown', $colour = 'not known' )
{
    $output = "$user is from {$country} and {$age} years old. His favourite colour is {$colour}.";

    // Only print the output if the user is 'Rob'
    if ( so_user_test( $user ) )
        return print $output;

    return print 'We are not allowed to give you information about this user.';
}

你可以这样调用它：

so_user_data( 'Fancy Franzy' );
// outputs: We are not allowed to give you information about this user.

so_user_data( 'Rob' );
// outputs: Rob is from an unknown country and unknown years old. His favourite colour is not known.

so_user_data( 'Silent Bob', 'U.S.A.', '38', 'brown' );
// outputs: Silent Bob is from U.S.A. and 38 years old. His favourite colour is brown.