Variable name as variable

问题:

I have a question about using variables as variablename. I have an array with database fieldnames as key and object properties as values like this:

$properties = array("userid" => "user['userid']", "city" => "hometown");  foreach ($properties as $field => $property ) {    $value1 = $db->$field;   $value2 = $obj->$property; } 

This works for the property hometown but doesn't work voor de property user['userid']. What is the correct way to adress the property variable?
I also tried several things like: ${property} or {$property} but without luck yet.

edit: Thanks for all the responses! For now I'll stay with my original solution, I was wondering if there was a way, I don't have principal problems with the eval version, will keep it in mind!

 foreach ($fields as $field => $property ) {         switch ($field) {             case "userid":                 $newvalue = $this->user['userid'];                 $oldvalue = $original->user['userid'];                 break;             // more cases ...             default:                 $newvalue = $this->{$property};                 $oldvalue = $original->($property};         }          .... 

回答1:

OK, so I decided to comment on this problem. Kai (no offence :) provided you a solution which doesn't use eval, but at the cost of additional 12 or so lines of code, making the code no doubt slower and more complex. So I myself think that in this case, it is justified to use eval() call. Hovewer, that could be just me (I value clear and short code a lot).

But beware, if input is coming from external sources, then you should filter it.

$properties = array("userid" => "user['userid']", "city" => "hometown");  foreach ($properties as $field => $property ) {    $value1 = $db->$field;   eval("\$value2=\$obj->$property;"); } 

回答2:

if $obj is an object and has a field $property

$obj->{$property} should be working fine

also you can use the curly brackets with concatenated strings like:

$obj->{ "field_" . $field_name }; 

In your case $property is a string, so on the first iteration it will be user['userid']

edit: to make it 2 dimensional array $properties have to be defined like this:

$properties = array(     'user' => array(         'id' => 1,         'name' => 'username'     ),     'city' => 'hometown' ); 

回答3:

Your problem is that you have an array index there and this is not supported by the "variable variables" syntax of PHP.

I know this is kind of magic, but you can do this without the use of evil eval().

Please note: this is just a proof of concept to show that it's possible.

What you really should to is refactor your code in a way so that such a hack is not necessary.

$bar = "hallo"; $foo['bar']['baz'] = "hallo2";  $properties = array('bar', "foo['bar']['baz']");  // DOES NOT WORK foreach ($properties as $property)         echo "$property = ", $$property, "\n";  /* Results in:  * bar = hallo  * foo['bar']['baz'] = PHP Notice:  Undefined variable: foo['bar']['baz'] in /tmp/test.php on line 8  */  // DOES WORK foreach ($properties as $property)         echo "$property = ", get_var($property), "\n";  /* Results in:  * bar = hallo  * foo['bar']['baz'] = hallo2  */  // dark magic starts here function get_var($name) {         if (strpos($name, '[') === false) {                 global $$name;                 return $$name;         } else {                 // split variable name into array name and nested index segments                 preg_match_all("#[^\[\]\"']+#", $name, $parts);                 $parts = $parts[0];                  // get pointer to array and walk down to the desired (nested) index                 $varname = array_shift($parts);                 global $$varname;                 $pointer =& $$varname;                 foreach ($parts as $index) {                         $pointer =& $pointer[$index];                 }                 return $pointer;         } } 

文章来源: Variable name as variable