타입 저글링
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.
변수
Table of Contents
기본문법
PHP에서 변수는 변수명 앞에 달러사인을 덧붙여 표현된다. 변수명은 대소문자를 구별한다.
PHP에서 변수명은 다음 규칙을 따른다. 유효한 변수명은 문자나 밑줄로 시작하고, 그 뒤에 문자, 숫자, 밑줄이 붙을수 있다. 정규표현식으로 표현하면 다음과 같다: '[a-zA-Z_\x7f-\xff][a-zA-Z0-9_\x7f-\xff]*'
Note: 여기서, 문자는 a-z, A-Z, 그리고 아스키문자(ASCII) 127에서 255까지(0x7f-0xff).
<?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 ASCII 228.
?>
PHP 3에서 변수는 항상 값에 의해 지정되야 한다(assinged by value). 변수를 표현식으로 지정할때에 원래 표현식의 모든 값이 목표 변수로 복사된다. 이 말의 의미는 예를 들면, 어떤 변수값을 다른 변수로 지정한 후에, 그 변수중 어떤 하나를 변경하는것이 다른 변수에 영향을 미치지 않는다는 의미를 갖는다. 이런 종류의 지정에 대해서 표현식을 참고.
PHP 4에서는 이와 다른 방법으로 변수에 값이 지정된다: 참조에 의한 지정. 이 용어의 의미는 새로운 변수가 원래 변수를 참조한다는 것이다.(즉, "원래 변수의 별명이 되는것" 이나 "가리키는 것") 새 변수의 변경은 원래 변수에 영향을 미치고, 그 반대도 가능하다. 이것은 복사가 수행되지 않는다는 것을 의미한다. 따라서, 지정은 매우 신속하게 이루어진다. 하지만, 속도향상은 견고한 루프나 거대한 배열이나 객체를 지정할때만 효과가 있을것이다.
참조에 의한 지정을 위해서는, 단순히 지정되는(소스 변수) 변수의 시작부분에 엠퍼센트(&)를 덧붙이면 된다. 예를 들면 다음 코드 예는 'My name is Bob'이 두번 출력된다.
<?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.
?>
주의할 것은 오직 이름이 부여된 변수만이 참조에 의해 지정된다는 것이다.
<?php
$foo = 25;
$bar = &$foo; // This is a valid assignment.
$bar = &(24 * 7); // Invalid; references an unnamed expression.
function test()
{
return 25;
}
$bar = &test(); // Invalid.
?>
add a note
User Contributed Notes변수
(rot13) yrvsrevx [at] jro [dot] qr
15-Feb-2008 07:00
15-Feb-2008 07:00
(rot13) yrvsrevx [at] jro [dot] qr
15-Feb-2008 04:22
15-Feb-2008 04:22
@cgorbit: This does not work, because "=" has a lower operator precedence than "||". The expression "($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 $key;
}
cgorbit
05-Jan-2008 03:34
05-Jan-2008 03:34
<?php
function var_name(&$var, $scope=0)
{
$old = $var;
if (($key = array_search($var = 'unique'.rand().'value', !$scope ? $GLOBALS : $scope)) && ($var = $old || true)) return $key;
}
?>
Because $var may casts to false
alexandre at nospam dot gaigalas dot net
06-Jul-2007 09:13
06-Jul-2007 09:13
Here's a simple solution for retrieving the variable name, based on the lucas (http://www.php.net/manual/en/language.variables.php#49997) solution, but shorter, just two lines =)
<?php
function var_name(&$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 08:48
20-Feb-2007 08:48
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 $title_default_Dr value="Dr">Dr</option>
</select>
END;
?>
code at slater dot fr
25-Jan-2007 02:10
25-Jan-2007 02:10
Here's a pair of functions to encode/decode any string to be a valid php and javascript variable name.
<?php
function label_encode($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 10:14
28-Dec-2006 10:14
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 01:44
04-Aug-2006 01:44
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 05:44
20-May-2006 05:44
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
27-Dec-2005 11:11
27-Dec-2005 11:11
> 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 02:03
25-Nov-2005 02:03
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 01:25
10-Nov-2005 01:25
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 04:33
13-Oct-2005 04:33
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 05:09
31-Aug-2005 05:09
Variables can also be assigned together.
<?php
$a = $b = $c = 1;
echo $a.$b.$c;
?>
This outputs 111.
Mike Fotes
09-Jul-2005 11:46
09-Jul-2005 11:46
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 01:06
17-May-2005 01:06
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 05:17
01-May-2005 05:17
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 04:01
25-Apr-2005 04:01
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 09:18
07-Apr-2005 09:18
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 04:42
14-Feb-2005 04:42
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
04-Feb-2005 11:45
04-Feb-2005 11:45
$id = 2;
$cube_2 = "Test";
echo ${cube_.$id};
// will output: Test
ringo78 at xs4all dot nl
14-Jan-2005 12:27
14-Jan-2005 12:27
<?
// 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 03:02
07-Jan-2005 03:02
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 10:58
25-May-2004 10:58
<?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 12:31
09-Mar-2004 12:31
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 08:15
20-Jan-2004 08:15
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 06:37
14-Jan-2003 06:37
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