假设我有这样的PHP代码:

$FooBar = "a string";

然后我需要一个这样的函数:

print_var_name($FooBar);

打印:

FooBar

有什么想法如何实现这一点?这在PHP中可能吗?


当前回答

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'];

其他回答

出于调试的原因,我做了一个检查函数。它就像print_r()的类固醇,很像Krumo,但对对象更有效一点。我想添加var名称检测,灵感来自于Nick Presta的帖子。它检测作为参数传递的任何表达式,而不仅仅是变量名。

这只是检测传递表达式的包装器函数。 大多数案子都没问题。 如果在同一行代码中多次调用该函数,则它将不起作用。

这很好: 死(检查($ this - > getUser()——> hasCredential(“删除”)));

Inspect()是检测传递表达式的函数。

我们得到:$this->getUser()->hasCredential("delete")

function inspect($label, $value = "__undefin_e_d__")
{
    if($value == "__undefin_e_d__") {

        /* The first argument is not the label but the 
           variable to inspect itself, so we need a label.
           Let's try to find out it's name by peeking at 
           the source code. 
        */

        /* The reason for using an exotic string like 
           "__undefin_e_d__" instead of NULL here is that 
           inspected variables can also be NULL and I want 
           to inspect them anyway.
        */

        $value = $label;

        $bt = debug_backtrace();
        $src = file($bt[0]["file"]);
        $line = $src[ $bt[0]['line'] - 1 ];

        // let's match the function call and the last closing bracket
        preg_match( "#inspect\((.+)\)#", $line, $match );

        /* let's count brackets to see how many of them actually belongs 
           to the var name
           Eg:   die(inspect($this->getUser()->hasCredential("delete")));
                  We want:   $this->getUser()->hasCredential("delete")
        */
        $max = strlen($match[1]);
        $varname = "";
        $c = 0;
        for($i = 0; $i < $max; $i++){
            if(     $match[1]{$i} == "(" ) $c++;
            elseif( $match[1]{$i} == ")" ) $c--;
            if($c < 0) break;
            $varname .=  $match[1]{$i};
        }
        $label = $varname;
    }

    // $label now holds the name of the passed variable ($ included)
    // Eg:   inspect($hello) 
    //             => $label = "$hello"
    // or the whole expression evaluated
    // Eg:   inspect($this->getUser()->hasCredential("delete"))
    //             => $label = "$this->getUser()->hasCredential(\"delete\")"

    // now the actual function call to the inspector method, 
    // passing the var name as the label:

      // return dInspect::dump($label, $val);
         // UPDATE: I commented this line because people got confused about 
         // the dInspect class, wich has nothing to do with the issue here.

    echo("The label is: ".$label);
    echo("The value is: ".$value);

}

下面是inspector函数(和我的dInspect类)的一个例子:

http://inspect.ip1.cc

该页面的文本是西班牙语,但代码简洁,非常容易理解。

我也想不出有效的方法但我想到了这个。对于下面的有限用途,它是有效的。

耸耸肩

<?php

function varName( $v ) {
    $trace = debug_backtrace();
    $vLine = file( __FILE__ );
    $fLine = $vLine[ $trace[0]['line'] - 1 ];
    preg_match( "#\\$(\w+)#", $fLine, $match );
    print_r( $match );
}

$foo = "knight";
$bar = array( 1, 2, 3 );
$baz = 12345;

varName( $foo );
varName( $bar );
varName( $baz );

?>

// Returns
Array
(
    [0] => $foo
    [1] => foo
)
Array
(
    [0] => $bar
    [1] => bar
)
Array
(
    [0] => $baz
    [1] => baz
)

它基于调用函数的行来工作,在那里它可以找到传入的参数。我认为它可以扩展到多个参数,但是,就像其他人说的,如果你能更好地解释情况,另一个解决方案可能会更好。

如果变量是可交换的,那么必须在某个地方有逻辑来决定使用哪个变量。你所需要做的就是在你做其他事情的时候,把变量名放在这个逻辑中的$variable中。

我想我们都很难理解你需要这个做什么。示例代码或对实际要做的事情的解释可能会有所帮助,但我怀疑您想得太多了。

我实际上有一个有效的用例。

我有一个函数cacheVariable($var)(好吧,我有一个函数缓存($key, $value),但我想有一个函数如上所述)。

目的是:

$colour = 'blue';
cacheVariable($colour);

...

// another session

...

$myColour = getCachedVariable('colour');

我试过了

function cacheVariable($variable) {
   $key = ${$variable}; // This doesn't help! It only gives 'variable'.
   // do some caching using suitable backend such as apc, memcache or ramdisk
}

我也试过

function varName(&$var) {
   $definedVariables = get_defined_vars();
   $copyOfDefinedVariables = array();
   foreach ($definedVariables as $variable=>$value) {
      $copyOfDefinedVariables[$variable] = $value;
   }
   $oldVar = $var;
   $var = !$var;
   $difference = array_diff_assoc($definedVariables, $copyOfDefinedVariables);
   $var = $oldVar;
   return key(array_slice($difference, 0, 1, true));
}

但这也失败了……:(

当然,我可以继续做缓存(' color ', $ color),但我很懒,你知道…;)

所以,我想要的是一个函数,它得到一个变量的原始名称,因为它被传递给一个函数。在函数内部,我不可能知道这一点。在上面的第二个例子中,通过引用传递get_defined_vars()在一定程度上帮助了我(感谢Jean-Jacques Guegan的这个想法)。后一个函数开始工作,但它仍然只返回局部变量('variable',而不是' color ')。

我还没有尝试使用get_func_args()和get_func_arg(), ${}-构造和key()组合,但我认为它也会失败。

PHP中没有可以输出变量名的预定义函数。但是,您可以使用get_defined_vars()的结果,该结果返回范围内定义的所有变量,包括名称和值。这里有一个例子:

<?php
    // Function for determining the name of a variable
    function getVarName(&$var, $definedVars=null) {
        $definedVars = (!is_array($definedVars) ? $GLOBALS : $definedVars);
        $val = $var;
        $rand = 1;
        while (in_array($rand, $definedVars, true)) {
            $rand = md5(mt_rand(10000, 1000000));
        }
        $var = $rand;
         
        foreach ($definedVars as $dvName=>$dvVal) {
            if ($dvVal === $rand) {
                $var = $val;
                return $dvName;
            }
        }
         
        return null;
    }
 
    // the name of $a is to be determined.   
    $a = 1;
     
    // Determine the name of $a
    echo getVarName($a);
?>

阅读更多在如何获得一个变量名作为一个字符串在PHP?