php - 如何将参数传递给函数

问题描述：

I'm pretty sure this is a very basic question to all of you, but i'm new with php, and i don't really get it... basically i've created a function in which i need to pass two parameters.

My functions is this:

function displayRoomDetails($customerRooms, $test)
{
    foreach ($customerRooms as $room) {
        $test.= $room->name;
    };
}

it is a very basic function, but will do for this example.

Now, i'm creating templates to display several information, and i have 3 different layout where i need to display the same info but styled differently, so my approach was:

template1 .= '<span>';

if (!$customerRooms == "") {
    displayRoomDetails($customerRooms,"template1");
};

template1 .= '</span>';

It should be pretty easy to understand, basically i'm calling the same functions in all the different templates passing as a parameter the template name and trying to append the results to the right template.

The problem i've got is this: According to this example here -> http://www.w3schools.com/php/showphp.asp?filename=demo_function3

i should be able to do this exactly like i did, but when i try, when i debug my function, $template doesn't take the passed value as i though it would, but it is still called $test and not $template1...

What am i doing wrong?

Thanks

我很确定这对你们所有人来说都是一个非常基本的问题，但我是php新手，我真的没有得到它... 基本上我已经创建了一个函数，我需要传递两个参数。 p>

我的函数是这样的： p>

  function displayRoomDetails（$ customerRooms，$ test）
 {
 foreach（$ customerRooms as $ room）{
 $ test。= $ room-＆gt; name; 
}; 
  } 
  code>  pre> 
 
 这是一个非常基本的功能，但是这个例子也适用。 p> 
 
 
现在，我正在创建模板 显示几个信息，我有3种不同的布局，我需要显示相同的信息，但风格不同，所以我的方法是： p> 
 
 
  template1。='＆lt; span＆gt;  '; 
 
if（！$ customerRooms ==“”）{
 displayRoomDetails（$ customerRooms，“template1”）; 
}; 
 
 
mpmplate1。='＆lt; / span＆gt;'; 
  代码>  pre> 
 
 它应该很容易理解，基本上我在所有传递的不同模板中调用相同的函数 一个参数模板名称，并尝试将结果附加到正确的模板。 p> 
 
 
我遇到的问题是：
根据这个例子这里 - ＆gt; 
  http://www.w3schools.com/php/showphp.asp?filename=demo_function3  a  >  p> 
 
 
我应该能够像我一样完成此操作，但是当我尝试时，当我调试我的函数时，$ template不会像我那样接受传递的值， 但它仍然被称为 $ test  code>而不是 $ template1 ...  code>  p> 
 
 
我做错了什么？ p> \  n 
 

谢谢 p> 
  div>

答

Try these changes:

function displayRoomDetails($customerRooms, &$test)

And

$template1 .= '<span>';
if ($customerRooms != "") {
  displayRoomDetails($customerRooms, $template1);
};
$template1 .= '</span>';

答

From what I understand you want to append some text to the template1 variable using the displayRoomDetailsFunction

Some things to fix:

template1 should be $template1
You should be passing the $template1 not the "template1" (i.e. the variable itself not its name).
If you want to modify this variable you need to either:
- pass it as reference, which you can do by changing the function's declaration to: function displayRoomDetails($customerRooms, &$test)
- return new string from function and assign it to the $template1 by adding return $test; just after your foreach block and changing the call to $template1 .= displayRoomDetails($customerRooms,$template1);

Additional note: if $customerRooms is an array, it'd be better to check if it's not empty using count() than !$customerRooms == "", see @andrew's comment for details

php - 如何将参数传递给函数

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