似乎没有人提到这是a)困难和b)不明智的根本原因:

A "variable" is just a symbol pointing at something else. In PHP, it internally points to something called a "zval", which can actually be used for multiple variables simultaneously, either because they have the same value (PHP implements something called "copy-on-write" so that $foo = $bar doesn't need to allocate extra memory straight away) or because they have been assigned (or passed to a function) by reference (e.g. $foo =& $bar). So a zval has no name. When you pass a parameter to a function you are creating a new variable (even if it's a reference). You could pass something anonymous, like "hello", but once inside your function, it's whatever variable you name it as. This is fairly fundamental to code separation: if a function relied on what a variable used to be called, it would be more like a goto than a properly separate function. Global variables are generally considered a bad idea. A lot of the examples here assume that the variable you want to "reflect" can be found in $GLOBALS, but this will only be true if you've structured your code badly and variables aren't scoped to some function or object. Variable names are there to help programmers read their code. Renaming variables to better suit their purpose is a very common refactoring practice, and the whole point is that it doesn't make any difference.

现在，我理解这种调试的愿望(尽管一些建议的用法远远超出了这一点)，但作为一种通用的解决方案，它实际上并没有你想象的那么有用:如果你的调试函数说你的变量是“$file”，那仍然可能是你代码中数十个“$file”变量中的任何一个，或者一个你称为“$filename”的变量，但传递给一个参数为“$file”的函数。

更有用的信息是在代码中调用调试函数的位置。因为你可以在你的编辑器中快速找到它，你可以看到你为自己输出的变量，甚至可以一次性将整个表达式传递给它(例如debug('$foo + $bar = ')。($foo + $bar))。

为此，你可以在调试函数的顶部使用这段代码:

$backtrace = debug_backtrace();
echo '# Debug function called from ' . $backtrace[0]['file'] . ' at line ' . $backtrace[0]['line'];

从php.net

@Alexandre -简短的解决方案

<?php
function vname(&$var, $scope=0)
{
    $old = $var;
    if (($key = array_search($var = 'unique'.rand().'value', !$scope ? $GLOBALS : $scope)) && $var = $old) return $key;  
}
?>

@Lucas - usage

<?php
//1.  Use of a variable contained in the global scope (default):
  $my_global_variable = "My global string.";
  echo vname($my_global_variable); // Outputs:  my_global_variable

//2.  Use of a local variable:
  function my_local_func()
  {
    $my_local_variable = "My local string.";
    return vname($my_local_variable, get_defined_vars());
  }
  echo my_local_func(); // Outputs: my_local_variable

//3.  Use of an object property:
  class myclass
  {
    public function __constructor()
    {
      $this->my_object_property = "My object property  string.";
    }
  }
  $obj = new myclass;
  echo vname($obj->my_object_property, $obj); // Outputs: my_object_property
?>

您可能会考虑改变您的方法，并使用一个变量变量名?

$var_name = "FooBar";
$$var_name = "a string";

然后你就可以

print($var_name);

得到

FooBar

这里是PHP变量手册的链接

我知道这个问题很老了，而且已经有人回答了，但我其实是在找这个。我把这个答案贴出来是为了给大家节省一点时间来完善一些答案。

选项1:

$data = array('$FooBar');  

$vars = [];  
$vars = preg_replace('/^\\$/', '', $data); 

$varname = key(compact($vars));  
echo $varname;

打印:

FooBar

不管出于什么原因，你会发现自己处于这样的情况下，它确实有效。

． 选项2:

$FooBar = "a string";  

$varname = trim(array_search($FooBar, $GLOBALS), " \t.");  
echo $varname;

如果$FooBar拥有一个唯一的值，它将打印'FooBar'。如果$FooBar为空或空，它将打印找到的第一个空字符串或空字符串的名称。

它可以这样使用:

if (isset($FooBar) && !is_null($FooBar) && !empty($FooBar)) {
    $FooBar = "a string";
    $varname = trim(array_search($FooBar, $GLOBALS), " \t.");
}

这是我基于杰里米·鲁滕的解决方案

class DebugHelper {

    function printVarNames($systemDefinedVars, $varNames) {
        foreach ($systemDefinedVars as $var=>$value) {
            if (in_array($var, $varNames )) {
                var_dump($var);
                var_dump($value);
            }
        }
    }
}

使用它

DebugHelper::printVarNames(
    $systemDefinedVars = get_defined_vars(),
    $varNames=array('yourVar00', 'yourVar01')
);

您可以使用get_defined_vars()来查找与您试图查找的变量名称具有相同值的变量的名称。显然这并不总是有效，因为不同的变量通常有相同的值，但这是我能想到的唯一方法。

编辑:get_defined_vars()似乎没有正确工作，它返回'var'，因为$var在函数本身中使用。$GLOBALS似乎可以，所以我把它改成了这个。

function print_var_name($var) {
    foreach($GLOBALS as $var_name => $value) {
        if ($value === $var) {
            return $var_name;
        }
    }

    return false;
}

编辑:需要明确的是，在PHP中没有好的方法来做到这一点，这可能是因为您不应该这样做。也许有更好的方法来做你想做的事情。