我的目标是回显传递给函数的参数。例如，如何做到这一点？
$contact_name = 'foo';
function do_something($some_argument){
// echo 'contact_name' .... How???
}
do_something($contact_name);
【问题讨论】：
Sander 有正确的答案，但这正是我正在寻找的东西：
$contact_name = 'foo';
function do_something($args = array(), $another_arg) {
foreach ($args as $name => $value) {
echo $name;
echo '<br>'.$another_arg;
}
}
do_something(compact(contact_name),'bar');
【讨论】：
基于 PTBNL 的（绝对正确）的答案，我想出了一个更具可读性（至少我认为如此）的方法：
/**
* returns the name of the variable posted as the first parameter.
* If not called from global scope, pass in get_defined_vars() as the second parameter
*
* behind the scenes:
*
* this function only works because we are passing the first argument by reference.
* 1. we store the old value in a known variable
* 2. we overwrite the argument with a known randomized hash value
* 3. we loop through the scope's symbol table until we find the known value
* 4. we restore the arguments original value and
* 5. we return the name of the symbol we found in the table
*/
function variable_name( & $var, array $scope = null )
{
if ( $scope == null )
{
$scope = $GLOBALS;
}
$__variable_name_original_value = $var;
$__variable_name_temporary_value = md5( number_format( microtime( true ), 10, '', '' ).rand() );
$var = $__variable_name_temporary_value;
foreach( $scope as $variable => $value )
{
if ( $value == $__variable_name_temporary_value && $variable != '__variable_name_original_value' )
{
$var = $__variable_name_original_value;
return $variable;
}
}
return null;
}
// prove that it works:
$test = 1;
$hello = 1;
$world = 2;
$foo = 100;
$bar = 10;
$awesome = 1;
function test_from_local_scope()
{
$local_test = 1;
$local_hello = 1;
$local_world = 2;
$local_foo = 100;
$local_bar = 10;
$local_awesome = 1;
return variable_name( $local_awesome, get_defined_vars() );
}
printf( "%s\n", variable_name( $awesome, get_defined_vars() ) ); // will echo 'awesome'
printf( "%s\n", test_from_local_scope() ); // will also echo awesome;
【讨论】：
直接离开 PHP.net variables 页面：
<?php
function vname(&$var, $scope=false, $prefix='unique', $suffix='value')
{
if($scope) $vals = $scope;
else $vals = $GLOBALS;
$old = $var;
$var = $new = $prefix.rand().$suffix;
$vname = FALSE;
foreach($vals as $key => $val) {
if($val === $new) $vname = $key;
}
$var = $old;
return $vname;
}
?>
【讨论】：
免责声明：仅当您将变量而不是值传递给函数时才有效，并且仅当您不在函数或类中时才有效。所以只有 GLOBAL 范围有效:)
Good funct($var)
Bad funct(1)
你可以做到与普遍认为的相反^_^。但它涉及到一些 $GLOBALS 变量的查找技巧。
你这样做：
$variable_name = "some value, better if its unique";
function funct($var) {
foreach ($GLOBALS as $name => $value) {
if ($value == $var) {
echo $name; // will echo variable_name
break;
}
}
}
这种方法不是万无一失的。因为如果两个变量具有相同的值，该函数将获取它找到的第一个变量的名称。不是你想要的：P 如果您想要变量名称的准确性，最好事先使变量值唯一
另一种方法是像这样使用参考来准确
$variable_name = 123;
function funct(&$var) {
$old = $var;
$var = $checksum = md5(time()); // give it unique value
foreach ($GLOBALS as $name => $value) {
if ($value == $var) {
echo $name; // will echo variable_name
$var = $old; // reassign old value
break;
}
}
}
所以完全有可能:)
【讨论】：
class Someone{
protected $name='';
public function __construct($name){
$this->name=$name;
}
public function doSomthing($arg){
echo "My name is: {$this->name} and I do {$arg}";
}
}
//in main
$Me=new Someone('Itay Moav');
$Me->doSomething('test');
【讨论】：
变量只是用于寻址内存中的值或区域的方法。您无法获取已将值传递给函数的变量名称。
【讨论】：
你不能。如果你想这样做，你还需要传递名称，例如：
$contact_name = 'foo';
$contact_phone = '555-1234';
function do_something($args = array()) {
foreach ($args as $name => $value) {
echo "$name: $value<br />";
}
}
do_something(compact('contact_name', 'contact_phone'));
【讨论】：
不可能。
【讨论】：