I get the basics of variable variables, but I saw a syntax just know, which bogles my mind a bit.
$this->{$toShow}();
I don't really see what those {} symbols are doing there. Do they have any special meaning?
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4 Answers
PHP's variable parser isn't greedy. The {} are used to indicate what should be considered part of a variable reference and what isn't. Consider this:
$arr = array();
$arr[3] = array();
$arr[3][4] = 'Hi there';
echo "$arr[3][4]";
Notice the double quotes. You'd expect this to output Hi there, but you actually end up seeing Array[4]. This is due to the non-greediness of the parser. It will check for only ONE level of array indexing while interpolating variables into the string, so what it really saw was this:
echo $arr[3], "[4]";
But, doing
echo "{$arr[3][4]}";
forces PHP to treat everything inside the braces as a variable reference, and you end up with the expected Hi there.
Comments
They tell the parser, where a variable name starts and ends. In this particular case it might not be needed, but consider this example:
$this->$toShow[0]
What should the parser do? Is $toShow an array or $this->$toShow ? In this case, the variable is resolved first and the array index is applied to the resulting property.
So if you actually want to access $toShow[0], you have to write:
$this->{$toShow[0]}
Comments
These curly braces can be used to use expressions to specify the variable identifier instead of just a variable’s value:
$var = 'foo';
echo ${$var.'bar'}; // echoes the value of $foobar
echo $$var.'bar'; // echoes the value of $foo concatenated with "bar"
Comments
$this->{$toShow}();
Break it down as below:
First this is a object oriented programming style as you got to see $this and ->. Second, {$toShow}() is a method(function) as you can see the () brackets.
So now {$toShow}() should somehow be parsed to a name like 'compute()'. And, $toShow is just a variable which might hold a possible function name. But the question remains, why is {} used around.
The reason is {} brackets substitues the value in the place. To clarify,
$toShow="compute";
so,
{$toShow}(); //is equivalent to compute();
but this is not true:
$toShow(); //this is wrong as a variablename is not a legal function name
