How to get variable name in PHP?

Question

func($name1) should return name1

Is it possible?

Answer 1

Here is the function that does this.

 function var_name (&$iVar, &$aDefinedVars) { foreach ($aDefinedVars as $k=>$v) $aDefinedVars_0[$k] = $v; $iVarSave = $iVar; $iVar =!$iVar; $aDiffKeys = array_keys (array_diff_assoc ($aDefinedVars_0, $aDefinedVars)); $iVar = $iVarSave; return $aDiffKeys[0]; } 

Call it like this:

 $test = "blah"; echo var_name($test, get_defined_vars()); 

This will print a "test."

I initially discovered that there is a function . You can also do this by iterating over the array returned by get_defined_vars (). This may be a little easier to understand.

Answer 2

No, there is no way to get the variable name in PHP.

When a function is called, this function receives only the contents of the variable, and not "the variable itself" - this means that the function cannot recognize the name of the variable that was passed to it.

Answer 3

Not.

Answer 4

A good idea? Not

Do you want you to do this? Not

Proof of concept? Of course!

 <?php a($test); function a($x) { $trace = debug_backtrace(); $file = file($trace[0]['file']); $line = $file[$trace[0]['line']-1]; var_dump($line); // Prints "a($test);" Do the Stringparsing and your done } 

Yes, it takes a “simple” reading of the source file, it can also be done using a php extension called “bytekit” that gives you user access to php opcodes and works from there.

Answer 5

Not.

When you define a function, you specify the local name of the variable so that it is inside the scope of that function. PHP will pass the corresponding value to the function, but the character is no longer in scope.

Alternatively, you could use " variable variables ".

Answer 6

Obviously, with sufficiently large values ​​of madness is possible.

The comments on this page include several methods: http://php.net/manual/en/language.variables.php

lucas dot karisny in linuxmail dot org answers my machine: http://www.php.net/manual/en/language.variables.php#49997

YMMV.