User Contributed Notes

Kubo2 22-Mar-2015 08:10
Note that if you use runtime variable name recognition in your code, you are able to use any string as a variable name. Consider following code:

<?php

$varName = 'foo with bar';
${'foo with bar'} = 42;

// will output int(42)
var_dump($$varName);

?>

This can be useful for example when accessing a property of an object constructed from JSON:

<?php

$composerJson = json_decode(file_get_contents(__DIR__ . '/composer.json'));

// would output sth. similar to: object(stdClass)#...
var_dump(
    $composerJson->{'require-dev'}
);

?>
justgook at gmail dot com 03-Aug-2010 10:38
I found interstate solution to work with arrays

<?php
$vars['product']['price']=11;

$aa='product';
$bb='price';

echo $vars{$aa}{$bb};

//prints 11
?>
dimitrov dot adrian at gmail dot com 29-Jul-2010 03:52
This is mine type casting lib, that is very useful for me.

<?php

function CAST_TO_INT($var, $min = FALSE, $max = FALSE)
{
    $var = is_int($var) ? $var : (int)(is_scalar($var) ? $var : 0);
    if ($min !== FALSE && $var < $min)
        return $min;
    elseif($max !== FALSE && $var > $max)
        return $max;
    return $var;
       
}

function CAST_TO_FLOAT($var, $min = FALSE, $max = FALSE)
{
    $var = is_float($var) ? $var : (float)(is_scalar($var) ? $var : 0);
    if ($min !== FALSE && $var < $min)
        return $min;
    elseif($max !== FALSE && $var > $max)
        return $max;
    return $var;
}

function CAST_TO_BOOL($var)
{
    return (bool)(is_bool($var) ? $var : is_scalar($var) ? $var : FALSE);
}

function CAST_TO_STRING($var, $length = FALSE)
{
    if ($length !== FALSE && is_int($length) && $length > 0)
        return substr(trim(is_string($var)
                    ? $var
                    : (is_scalar($var) ? $var : '')), 0, $length);

    return trim(
                is_string($var)
                ? $var
                : (is_scalar($var) ? $var : ''));
}

function CAST_TO_ARRAY($var)
{
    return is_array($var)
            ? $var
            : is_scalar($var) && $var
                ? array($var)
                : is_object($var) ? (array)$var : array();
}

function CAST_TO_OBJECT($var)
{
    return is_object($var)
            ? $var
            : is_scalar($var) && $var
                ? (object)$var
                : is_array($var) ? (object)$var : (object)NULL;
}

?>
Anonymous 20-Jul-2008 03:25
[EDIT by danbrown AT php DOT net: The function provided by this author will give you all defined variables at runtime.  It was originally written by (john DOT t DOT gold AT gmail DOT com), but contained some errors that were corrected in subsequent posts by (ned AT wgtech DOT com) and (taliesin AT gmail DOT com).]

<?php

echo '<table border=1><tr> <th>variable</th> <th>value</th> </tr>';
foreach( get_defined_vars() as $key => $value)
{
    if (is_array ($value) )
    {
        echo '<tr><td>$'.$key .'</td><td>';
        if ( sizeof($value)>0 )
        {
        echo '"<table border=1><tr> <th>key</th> <th>value</th> </tr>';
        foreach ($value as $skey => $svalue)
        {
            echo '<tr><td>[' . $skey .']</td><td>"'. $svalue .'"</td></tr>';
        }
        echo '</table>"';
        }
             else
        {
            echo 'EMPTY';
        }
        echo '</td></tr>';
    }
    else
    {
            echo '<tr><td>$' . $key .'</td><td>"'. $value .'"</td></tr>';
    }
}
echo '</table>';
?>
jsb17 at cornell dot edu 20-Feb-2007 05:48
As an addendum to David's 10-Nov-2005 posting, remember that curly braces literally mean "evaluate what's inside the curly braces" so, you can squeeze the variable variable creation into one line, like this:

<?php
  ${"title_default_" . $title} = "selected";
?>

and then, for example:

<?php
  $title_select = <<<END
    <select name="title">
      <option>Select</option>
      <option $title_default_Mr  value="Mr">Mr</option>
      <option $title_default_Ms  value="Ms">Ms</option>
      <option $title_default_Mrs value="Mrs">Mrs</option>
      <option $title_default_Dr  value="Dr">Dr</option>
    </select>
END;
?>
Mike at ImmortalSoFar dot com 25-Nov-2005 11:03
References and "return" can be flakey:

<?php
//  This only returns a copy, despite the dereferencing in the function definition
function &GetLogin ()
{
    return $_SESSION['Login'];
}

//  This gives a syntax error
function &GetLogin ()
{
    return &$_SESSION['Login'];
}

//  This works
function &GetLogin ()
{
    $ret = &$_SESSION['Login'];
    return $ret;
}
?>
david at removethisbit dot futuresbright dot com 10-Nov-2005 10:25
When using variable variables this is invalid:

<?php
$my_variable_{$type}_name = true;
?>

to get around this do something like:

<?php
$n="my_variable_{$type}_name";
${$n} = true;
?>

(or $$n - I tend to use curly brackets out of habit as it helps t reduce bugs ...)
Chris Hester 31-Aug-2005 02:09
Variables can also be assigned together.

<?php
$a = $b = $c = 1;
echo $a.$b.$c;
?>

This outputs 111.
josh at PraxisStudios dot com 17-May-2005 10:06
As with echo, you can define a variable like this:

<?php

$text = <<<END

<table>
    <tr>
        <td>
             $outputdata
        </td>
     </tr>
</table>

END;

?>

The closing END; must be on a line by itself (no whitespace).

[EDIT by danbrown AT php DOT net: This note illustrates HEREDOC syntax.  For more information on this and similar features, please read the "Strings" section of the manual here: http://www.php.net/manual/en/language.types.string.php ]
Carel Solomon 07-Jan-2005 12:02
You can also construct a variable name by concatenating two different variables, such as:

<?php

$arg = "foo";
$val = "bar";

//${$arg$val} = "in valid";     // Invalid
${$arg . $val} = "working";

echo $foobar;     // "working";
//echo $arg$val;         // Invalid
//echo ${$arg$val};     // Invalid
echo ${$arg . $val};    // "working"

?>

Carel
raja shahed at christine nothdurfter dot com 25-May-2004 07:58
<?php
error_reporting(E_ALL);

$name = "Christine_Nothdurfter";
// not Christine Nothdurfter
// you are not allowed to leave a space inside a variable name ;)
$$name = "'s students of Tyrolean language ";

print " $name{$$name}<br>";
print  "$name$Christine_Nothdurfter";
// same
?>
webmaster at daersys dot net 20-Jan-2004 05:15
You don't necessarily have to escape the dollar-sign before a variable if you want to output its name.

You can use single quotes instead of double quotes, too.

For instance:

<?php
$var = "test";

echo "$var"; // Will output the string "test"

echo "\$var"; // Will output the string "$var"

echo '$var'; // Will do the exact same thing as the previous line
?>

Why?
Well, the reason for this is that the PHP Parser will not attempt to parse strings encapsulated in single quotes (as opposed to strings within double quotes) and therefore outputs exactly what it's being fed with :)

To output the value of a variable within a single-quote-encapsulated string you'll have to use something along the lines of the following code:

<?php
$var = 'test';
/*
Using single quotes here seeing as I don't need the parser to actually parse the content of this variable but merely treat it as an ordinary string
*/

echo '$var = "' . $var . '"';
/*
Will output:
$var = "test"
*/
?>

HTH
- Daerion