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Predefined variables> <Type Juggling
Last updated: Fri, 11 Jul 2008

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Variables

Table of Contents

Basics

Variables in PHP are represented by a dollar sign followed by the name of the variable. The variable name is case-sensitive.

Variable names follow the same rules as other labels in PHP. A valid variable name starts with a letter or underscore, followed by any number of letters, numbers, or underscores. As a regular expression, it would be expressed thus: '[a-zA-Z_\x7f-\xff][a-zA-Z0-9_\x7f-\xff]*'

Note: For our purposes here, a letter is a-z, A-Z, and the ASCII characters from 127 through 255 (0x7f-0xff).

Note: $this is a special variable that can't be assigned.

Tip

See also the Userland Naming Guide.

For information on variable related functions, see the Variable Functions Reference.

<?php
$var 
'Bob';
$Var 'Joe';
echo 
"$var, $Var";      // outputs "Bob, Joe"

$4site 'not yet';     // invalid; starts with a number
$_4site 'not yet';    // valid; starts with an underscore
$täyte 'mansikka';    // valid; 'ä' is (Extended) ASCII 228.
?>

By default, variables are always assigned by value. That is to say, when you assign an expression to a variable, the entire value of the original expression is copied into the destination variable. This means, for instance, that after assigning one variable's value to another, changing one of those variables will have no effect on the other. For more information on this kind of assignment, see the chapter on Expressions.

PHP also offers another way to assign values to variables: assign by reference. This means that the new variable simply references (in other words, "becomes an alias for" or "points to") the original variable. Changes to the new variable affect the original, and vice versa.

To assign by reference, simply prepend an ampersand (&) to the beginning of the variable which is being assigned (the source variable). For instance, the following code snippet outputs 'My name is Bob' twice:

<?php
$foo 
'Bob';              // Assign the value 'Bob' to $foo
$bar = &$foo;              // Reference $foo via $bar.
$bar "My name is $bar";  // Alter $bar...
echo $bar;
echo 
$foo;                 // $foo is altered too.
?>

One important thing to note is that only named variables may be assigned by reference.

<?php
$foo 
25;
$bar = &$foo;      // This is a valid assignment.
$bar = &(24 7);  // Invalid; references an unnamed expression.

function test()
{
   return 
25;
}

$bar = &test();    // Invalid.
?>

It is not necessary to initialize variables in PHP however it is a very good practice. Uninitialized variables have a default value of their type - FALSE, zero, empty string or an empty array.

Example #1 Default values of uninitialized variables

<?php
echo ($unset_bool "true" "false"); // false
$unset_int += 25// 0 + 25 => 25
echo $unset_string "abc"// "" . "abc" => "abc"
$unset_array[3] = "def"// array() + array(3 => "def") => array(3 => "def")
?>

Relying on the default value of an uninitialized variable is problematic in the case of including one file into another which uses the same variable name. It is also a major security risk with register_globals turned on. E_NOTICE level error is issued in case of working with uninitialized variables, however not in the case of appending elements to the uninitialized array. isset() language construct can be used to detect if a variable has been already initialized.



Predefined variables> <Type Juggling
Last updated: Fri, 11 Jul 2008
 
add a note add a note User Contributed Notes
Variables
taliesin at gmail dot com
24-May-2008 11:14
or you could change the order of the if

function var_name(&$var, $scope=0)
{
    if (($old = $var) &&  (($var = $old) || true) && ($key = array_search($var = 'unique'.rand().'value', !$scope ? $GLOBALS : $scope))) return $key;
}
ned at wgtech dot com
22-May-2008 08:07
The function given in lower posts:

function var_name(&$var, $scope=0)
{
    if (($old = $var) && ($key = array_search($var = 'unique'.rand().'value', !$scope ? $GLOBALS : $scope)) && (($var = $old) || true)) return $key;
}

has a nasty problem. If array_search fails the (($var=$old)||true) never gets evaluated, and rightly so, permanently changing the value of $var to 'unique#####value'.

How to fail the function:

$z=1;
var_name($z,array(1));

Easily fixed by adding an else:

function var_name(&$var, $scope=false)
{
    if ((($old=$var)||true)&&($key = array_search($var = 'unique'.rand().'value', !$scope ? $GLOBALS : $scope)) && (($var = $old) || true)) return $key;
    else $var=$old;return false;
}

All better.
john dot t dot gold at gmail dot com
20-May-2008 06:03
To list all Variables for debug purposes use this:

<?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>"'. $value .'"</td></tr>';
    }
}
echo
'</table>';
?>
(rot13) yrvsrevx [at] jro [dot] qr
15-Feb-2008 04:00
err, the "($old = $var)" could also evaluate to false. Adding a "|| true" would make this overall to crude, I guess. It's better to leave it out of the expression.
(rot13) yrvsrevx [at] jro [dot] qr
15-Feb-2008 01:22
@cgorbit: This does not work, because "=" has a lower operator precedence than "||". The expression "($var = $old || true)" assigns true to $var. I corrected this and further shortened the function by putting "$old = $var" into the expression, too. :)
Note the parentheses that are necessary because of the operator precedence of "=" and "&&".

function var_name(&$var, $scope=0)
{
    if (($old = $var) && ($key = array_search($var = 'unique'.rand().'value', !$scope ? $GLOBALS : $scope)) && (($var = $old) || true)) return $key;
}
cgorbit
05-Jan-2008 12:34
<?php
function var_name(&$var, $scope=0)
{
   
$old = $var;
    if ((
$key = array_search($var = 'unique'.rand().'value', !$scope ? $GLOBALS : $scope)) && ($var = $old || true)) return $key
}
?>

Because $var may casts to false
alexandre at nospam dot gaigalas dot net
07-Jul-2007 06:13
Here's a simple solution for retrieving the variable name, based on the lucas (http://www.php.net/manual/en/language.variables.php#49997) solution, but shorter, just two lines =)

<?php
function var_name(&$var, $scope=0)
{
   
$old = $var;
    if ((
$key = array_search($var = 'unique'.rand().'value', !$scope ? $GLOBALS : $scope)) && $var = $old) return $key
}
?>
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;
?>
code at slater dot fr
25-Jan-2007 11:10
Here's a pair of functions to encode/decode any string to be a valid php and javascript variable name.

<?php

function label_encode($txt) {
 
 
// add Z to the begining to avoid that the resulting
  // label is a javascript keyword or it starts with a
  // number
 
$txt = 'Z'.$txt;
 
 
// encode as urlencoded data
 
$txt = rawurlencode($txt);
 
 
// replace illegal characters
 
$illegal = array('%', '-', '.');
 
$ok = array('é', 'è', 'à');
 
$txt = str_replace($illegal,$ok, $txt);
 
  return
$txt;
}

function
label_decode($txt) {
 
 
// replace illegal characters
 
$illegal = array('%', '-', '.');
 
$ok = array('é', 'è', 'à');
 
$txt = str_replace($ok, $illegal, $txt);
 
 
// unencode
 
$txt = rawurldecode($txt);
 
 
// remove the leading Z and return
 
return substr($txt,1);
}

?>
whoami
28-Dec-2006 07:14
what is so simple and flexible about these variable..? They're all the same thing -.-"
$var = whatever;

in fact is more complicated than:

String HelloWorld = hello;
giunta dot gaetano at sea-aeroportimilano dot it
04-Aug-2006 10:44
With php 5.1.4 (and maybe earlier?) take care about not using $this as a variable name, even when in the global scope or inside a plain function: the engine will prevent assigning any value to it...
molnaromatic at gmail dot com
20-May-2006 02:44
Simple sample and variables and html "templates":
The PHP code:
variables.php:
<?php
$SYSN
["title"] = "This is Magic!";
$SYSN["HEADLINE"] = "Ez magyarul van"; // This is hungarian
$SYSN["FEAR"] = "Bell in my heart";
?>

index.php:
<?php
include("variables.php");
include(
"template.html");
?>

The template:
template.html

<html>
<head><title><?=$SYSN["title"]?></title></head>
<body>
<H1><?=$SYSN["HEADLINE"]?></H1>
<p><?=$SYSN["FEAR"]?></p>
</body>
</html>
This is simple, quick and very flexibile
warhog at warhog dot net
28-Dec-2005 08:11
> Variable names follow the same rules as other labels in PHP. A valid variable name starts with a letter or underscore, followed by any number of letters, numbers, or underscores. As a regular expression, it would be expressed thus: '[a-zA-Z_\x7f-\xff][a-zA-Z0-9_\x7f-\xff]*'

..is not quite true. You can, in fact, only declare variables having a name like this if you use the syntax <?php $varname = "naks naks"; ?>.. but in fact a variable can have moreless any name that is a string... e.g. if you look at an array you can have
<?php
$arr
[''];
$arr['8'];
$arr['-my-element-is-so-pretty-useless-'];
?>
.. by accessing the variables-namespace via {} you can have the same functinalities for all variables, e.g.

<?php ${''} = "my empty variable"; ?>

is a valid expression and the variable having the empty string as name will have the value "my empty variable".

read the chapter on "variable variables" for further information.
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:

$my_variable_{$type}_name = true;

to get around this do something like:

$n="my_variable_{$type}_name";
${$n} = true;

(or $$n - I tend to use curly brackets out of habit as it helps t reduce bugs ...)
ludvig dot ericson at gmail dot com
13-Oct-2005 01:33
On the previous note:

This is due to how evaluation works. PHP will think of it as:

$a = whatever $b = $c is
$b = whatever $c = 1 is

... because an expression is equal to what it returns.

Therefore $c = 1 returns 1, making $b = $c same as $b = 1, which makes $b 1, which makes $a be $b, which is 1.

$a = ($b = $c = 1) + 2;
Will have $a be 3 while $b and $c is 1.

Hope that clears something up.
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.
Mike Fotes
09-Jul-2005 08:46
In conditional assignment of variables, be careful because the strings may take over the value of the variable if you do something like this:

<?php
$condition
= true;

// Outputs " <-- That should say test"
echo "test" . ($condition) ? " <-- That should say test" : "";
?>

You will need to enclose the conditional statement and assignments in parenthesis to have it work correctly:

<?php
$condition
= true;

// Outputs "test <-- That should say test"
echo "test" . (($condition) ? " <-- That should say test " : "");
?>
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).
user at host dot network
02-May-2005 02:17
pay attention using spaces, dots and parenthesis in case kinda like..
$var=($number>0)?1.'parse error':0.'here too';
the correct form is..
$var=($number>0)?1 .'parse error':0 .'here too';
or
$var=($number>0)?(1).'parse error':(0).'here too';
or
$var = ($number > 0) ? 1 . 'parse error' : 0 . 'here too';
etc..
i think that's why the parser read 1. and 0. like decimal numbers not correctly written, point of fact
$var=$number>0?1.0.'parse error':0.0.'here too';
seems to work correctly..
david at rayninfo dot co dot uk
26-Apr-2005 01:01
When constructing strings from text and variables you can use curly braces to "demarcate" variables from any surrounding text where, for whatever reason, you cannot use a space eg:

$str="Hi my name is ${bold}$name bla-bla";

which AFAIK is the same as

$str="Hi my name is {$bold}$name bla-bla";

zzapper
mike at go dot online dot pt
07-Apr-2005 06:18
In addition to what jospape at hotmail dot com and ringo78 at xs4all dot nl wrote, here's the sintax for arrays:

<?php
//considering 2 arrays
$foo1 = array ("a", "b", "c");
$foo2 = array ("d", "e", "f");

//and 2 variables that hold integers
$num = 1;
$cell = 2;

echo ${
foo.$num}[$cell]; // outputs "c"

$num = 2;
$cell = 0;

echo ${
foo.$num}[$cell]; // outputs "d"
?>
lucas dot karisny at linuxmail dot org
15-Feb-2005 01:42
Here's a function to get the name of a given variable.  Explanation and examples below.

<?php
 
function vname(&$var, $scope=false, $prefix='unique', $suffix='value')
  {
    if(
$scope) $vals = $scope;
    else     
$vals = $GLOBALS;
   
$old = $var;
   
$var = $new = $prefix.rand().$suffix;
   
$vname = FALSE;
    foreach(
$vals as $key => $val) {
      if(
$val === $new) $vname = $key;
    }
   
$var = $old;
    return
$vname;
  }
?>

Explanation:

The problem with figuring out what value is what key in that variables scope is that several variables might have the same value.  To remedy this, the variable is passed by reference and its value is then modified to a random value to make sure there will be a unique match.  Then we loop through the scope the variable is contained in and when there is a match of our modified value, we can grab the correct key.

Examples:

1.  Use of a variable contained in the global scope (default):
<?php
  $my_global_variable
= "My global string.";
  echo
vname($my_global_variable); // Outputs:  my_global_variable
?>

2.  Use of a local variable:
<?php
 
function my_local_func()
  {
   
$my_local_variable = "My local string.";
    return
vname($my_local_variable, get_defined_vars());
  }
  echo
my_local_func(); // Outputs: my_local_variable
?>

3.  Use of an object property:
<?php
 
class myclass
 
{
    public function
__constructor()
    {
     
$this->my_object_property = "My object property  string.";
    }
  }
 
$obj = new myclass;
  echo
vname($obj->my_object_property, $obj); // Outputs: my_object_property
?>
jospape at hotmail dot com
05-Feb-2005 08:45
$id = 2;
$cube_2 = "Test";

echo ${cube_.$id};

// will output: Test
ringo78 at xs4all dot nl
14-Jan-2005 09:27
<?
// I am beginning to like curly braces.
// I hope this helps for you work with them
$filename0="k";
$filename1="kl";
$filename2="klm";
 
$i=0;
for (
$varname = sprintf("filename%d",$i);   isset  ( ${$varname} ) ;   $varname = sprintf("filename%d", $i)  )  {
    echo
"${$varname} <br>";
   
$varname = sprintf("filename%d",$i);
   
$i++;
}
?>
Carel Solomon
07-Jan-2005 12:02
You can also construct a variable name by concatenating two different variables, such as:

<?

$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 surrealwebs dot com
09-Mar-2004 09:31
OK how about a practicle use for this:

You have a session variable such as:
$_SESSION["foo"] = "bar"
and you want to reference it to change it alot throughout the program instaed of typing the whole thing over and over just type this:

$sess =& $_SESSION
$sess['foo'] = bar;

echo $sess['foo'] // returns bar
echo $_SESSION["foo"] // also returns bar
just saves alot of time in the long run

also try $get = $HTTP_GET_VARS
or $post = $HTTP_POST_VARS
webmaster at daersys dot net
20-Jan-2004 05:15
In reference to "remco at clickbizz dot nl"'s note I would like to add that you don't necessarily have to escape the dollar-sign before a variable if you want to output it's 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
unleaded at nospam dot unleadedonline dot net
15-Jan-2003 03:37
References are great if you want to point to a variable which you don't quite know the value yet ;)

eg:

$error_msg = &$messages['login_error']; // Create a reference

$messages['login_error'] = 'test'; // Then later on set the referenced value

echo $error_msg; // echo the 'referenced value'

The output will be:

test

Predefined variables> <Type Juggling
Last updated: Fri, 11 Jul 2008
 
 
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