Chapter 12. Variables

Table of Contents
Basics
Predefined variables
Variable scope
Variable variables
Variables from outside PHP

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).

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.

?>

In PHP 3, 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.

As of PHP 4, PHP 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. This also means that no copying is performed; thus, the assignment happens more quickly. However, any speedup will likely be noticed only in tight loops or when assigning large arrays or objects.

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.

?>

User Contributed Notes
Variables

josh at PraxisStudios dot com
17-May-2005 03:06


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
01-May-2005 07:17


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
25-Apr-2005 06:01


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 11:18


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
14-Feb-2005 06:42


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 01:45


$id = 2;

$cube_2 = "Test";



echo ${cube_.$id};



// will output: Test

ringo78 at xs4all dot nl
14-Jan-2005 02:27


<?

// 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 05:02


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 12:58


<?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 02:31


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 10:15


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
14-Jan-2003 08:37


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