函数回声与返回

问题描述

好的，所以我有一些代码给我适合。

调用函数 d（$ z）;

<$>

$ z本身是代码中其他地方声明的变量。 p $ p> 函数d（$ x1）{
$ distance = $ x1;
返回$ distance;

$ / code>

这样做：



 函数d（$ x1）{
 $ distance = $ x1; 
 echo $ distance; 
} 
  

任何想法为何？

注意：我不想让这个值回显。我希望能够稍后使用该值作为变量，并且不一定会打印到屏幕上。 （现在，上面是一个简化版本，很多已被删除，我真正想做的事情，但即使上述不会正确返回给我）

虽然这不是我想要做的，但因为我想使用$ distance的结果作为内部变量（而不是回显它），所以如果我做了<？= $ distance？>< code>它不适用于返回。

解决方案

有两种可能的（可能更多）问题：

1. 您没有将正确的值传递给函数
   
   

   或

你没有将返回值赋给变量（例如 $ value = d（some value here）;




编辑：不要忘记echo $ value（存储返回）

>


Okay, so I have some code that is giving me fits.

To call the function d($z);

$z itself is a declared variable elsewhere in the code.

This doesn't work:

function d($x1){
$distance = $x1;
return $distance;
}

This works:

function d($x1){
$distance = $x1;
echo $distance;
}

Any ideas why?

Note: I don't want the value echo'd. I want to be able to use the value later on, as a variable, and not necessarily printed to the screen. (Now, the above is a simplified version, a lot has been removed, of what I really want to do, but even the above doesn't return properly for me)

Although it's not what I want to do, since I want to use the result of $distance as an internal variable (not echo it), if I do <?=$distance?> it doesn't work with the return.

解决方案

There are 2 possible (probably more) problems:

1. You are not passing a correct value into the function

   or

2. You are not assigning the return to a variable (ex. $value = d("some value here");)

Edit: Don't forget to echo $value (stored return)

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