Question

I’ve been working on a small set of classes in PHP 5.3.x that follow the Active Record pattern. I’m running to an issue, however, when working with static properties. Here’s the bit of code I’ve been testing this with:

<?php

class dbPreparedObject {

    public static $insert = "";

    public function __construct() {
        static::$insert = "autoinsert_".get_called_class();
    }
}

class gtRecord extends dbPreparedObject {}
class nRecord extends dbPreparedObject {}

$a = new gtRecord();
$b = new nRecord();

var_dump(gtRecord::$insert);
var_dump(nRecord::$insert);

Output:

string(18) "autoinsert_nRecord"
string(18) "autoinsert_nRecord"

I, however, expect the first string to read autoinsert_gtRecord.

It seems that static properties that aren’t instantiated by the child class are tied together. Is there any way to separate them without declaring public static $insert = "" in every child class?

Answer 1

This is the expected behavior. In PHP, properties declared static are shared amongst all instances of the object created. That is, there is only one instance of $insert across all instances of of dbPreparedObject or its inherited classes.

The reason you are seeing autoinsert_nRecord for both var_dumps is because that was the last object you created.

Consider this minor change:

$a = new gtRecord();
var_dump(gtRecord::$insert); // autoinsert_gtRecord

$b = new nRecord();    
var_dump(nRecord::$insert);  // autoinsert_nRecord
var_dump(gtRecord::$insert); // autoinsert_nRecord (because it was the last one set)

After the first var_dump, the value is _gtRecord, but once an nRecord object is created, the static property gets changed (for all dbPreparedObject objects) to be nRecord since that was the last class.

So if you plan on having multiple instances of this class, $insert cannot be static because it will not always contain the value you expect given the code since there is only one copy of the static $insert property that is the same for all objects. Once you change it in one object, you change it in all objects.

So we need to ask why $insert needs to be static or what other options to you have so you don’t run into this issue.