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?
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
为什么我们必须使用全局变量来获取变量名…我们可以像下面这样简单地使用。
$variableName = "ajaxmint";
echo getVarName('$variableName');
function getVarName($name) {
return str_replace('$','',$name);
}
其他回答
这正是你想要的-这是一个随时可以使用的“复制并导入”函数,它会回显给定变量的名称:
function print_var_name(){
// read backtrace
$bt = debug_backtrace();
// read file
$file = file($bt[0]['file']);
// select exact print_var_name($varname) line
$src = $file[$bt[0]['line']-1];
// search pattern
$pat = '#(.*)'.__FUNCTION__.' *?\( *?(.*) *?\)(.*)#i';
// extract $varname from match no 2
$var = preg_replace($pat, '$2', $src);
// print to browser
echo '<pre>' . trim($var) . ' = ' . print_r(current(func_get_args()), true) . '</pre>';
}
用法:print_var_name ($ FooBar)
打印:FooBar
提示 现在你可以重命名函数,它仍然可以工作,也可以在一行中多次使用该函数!感谢@Cliffordlife 我添加了一个更好的输出!多亏了@Blue-Water
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 ' . $backtrace[0]['line'];
从上面的答案改编为许多变量,具有良好的性能,只需一个$GLOBALS扫描许多
function compact_assoc(&$v1='__undefined__', &$v2='__undefined__',&$v3='__undefined__',&$v4='__undefined__',&$v5='__undefined__',&$v6='__undefined__',&$v7='__undefined__',&$v8='__undefined__',&$v9='__undefined__',&$v10='__undefined__',&$v11='__undefined__',&$v12='__undefined__',&$v13='__undefined__',&$v14='__undefined__',&$v15='__undefined__',&$v16='__undefined__',&$v17='__undefined__',&$v18='__undefined__',&$v19='__undefined__'
) {
$defined_vars=get_defined_vars();
$result=Array();
$reverse_key=Array();
$original_value=Array();
foreach( $defined_vars as $source_key => $source_value){
if($source_value==='__undefined__') break;
$original_value[$source_key]=$$source_key;
$new_test_value="PREFIX".rand()."SUFIX";
$reverse_key[$new_test_value]=$source_key;
$$source_key=$new_test_value;
}
foreach($GLOBALS as $key => &$value){
if( is_string($value) && isset($reverse_key[$value]) ) {
$result[$key]=&$value;
}
}
foreach( $original_value as $source_key => $original_value){
$$source_key=$original_value;
}
return $result;
}
$a = 'A';
$b = 'B';
$c = '999';
$myArray=Array ('id'=>'id123','name'=>'Foo');
print_r(compact_assoc($a,$b,$c,$myArray) );
//print
Array
(
[a] => A
[b] => B
[c] => 999
[myArray] => Array
(
[id] => id123
[name] => Foo
)
)
使用此方法将用户变量从全局变量分离到当前的检查变量。
function get_user_var_defined ()
{
return array_slice($GLOBALS,8,count($GLOBALS)-8);
}
function get_var_name ($var)
{
$vuser = get_user_var_defined();
foreach($vuser as $key=>$value)
{
if($var===$value) return $key ;
}
}
您可能会考虑改变您的方法,并使用一个变量变量名?
$var_name = "FooBar";
$$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