PHP Programer - 讨论区
标题：PHP的可变变量名
2010年11月18日 星期四 09:01
Variable variables
Sometimes it is convenient to be able to have variable variable names. That is, a variable name which can be set and used dynamically. A normal variable is set with a statement such as:
<?php
$a
=
'hello'
;
?>
A variable variable takes the value of a variable and treats that as the name of a variable. In the above example, hello , can be used as the name of a variable by using two dollar signs. i.e.
<?php
$
$a
=
'world'
;
?>
At this point two variables have been defined and stored in the PHP symbol tree: $a with contents "hello" and $hello with contents "world". Therefore, this statement:
<?php
echo
"
$a
${
$a
}
"
;
?>
produces the exact same output as:
<?php
echo
"
$a
$hello
"
;
?>
i.e. they both produce: hello world .
In order to use variable variables with arrays, you have to resolve an ambiguity problem. That is, if you write $$a[1] then the parser needs to know if you meant to use $a[1] as a variable, or if you wanted $$a as the variable and then the [1] index from that variable. The syntax for resolving this ambiguity is: ${$a[1]} for the first case and ${$a}[1] for the second.
Class properties may also be accessed using variable property names. The variable property name will be resolved within the scope from which the call is made. For instance, if you have an expression such as $foo->$bar , then the local scope will be examined for $bar and its value will be used as the name of the property of $foo . This is also true if $bar is an array access.
Example #1 Variable function example
<?php
class
foo
{
var
$bar
=
'I am bar.'
;
}
$foo
= new
foo
();
$bar
=
'bar'
;
$baz
= array(
'foo'
,
'bar'
,
'baz'
,
'quux'
);
echo
$foo
->
$bar
.
"\n"
;
echo
$foo
->
$baz
[
1
] .
"\n"
;
?>
The above example will output:
I am bar.
I am bar.
Please note that variable variables cannot be used with PHP's Superglobal arrays within functions or class methods. The variable $this is also a special variable that cannot be referenced dynamically.
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