型の相互変換
This script has been running around for a while and has add many additions and changes, however I noticed that if you run it and there is an object assigned to a variable it will fail with 'Object of Class 'XXXX' cannot be converted to string'
so I changed it a little to ignore objects within the script and display the variables only:
Btw, it has haelped me an ENOURMOUS amount over the years especially when keeping track of variables that are being sent through request and if session variables are alive
<?php
//this can be removed if you want to find out if you page has actually started a session, - this is sometimes a problem when relying on a session being started by an included/required file
if(!isset($_SESSION))
{
session_start();
}
//now begin the table
echo '<table border=1><tr> <th>variable</th> <th>value</th> </tr>';
foreach(get_defined_vars() as $key => $value)
{
//if the returned value is NOT an object...
if (is_array ($value) && !is_object($key) && !is_object($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 => $svalue )
{
//and if these values are NOT an object...
if ( !is_object($skey) && !is_object($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>';
?>
第 12章変数
基本的な事
PHP の変数はドル記号の後に変数名が続く形式で表されます。 変数名は大文字小文字を区別します。
変数名は、PHPの他のラベルと同じルールに従います。 有効な変数名は文字またはアンダースコアから始まり、任意の数の文字、 数字、アンダースコアが続きます。正規表現によれば、これは次の ように表現することができます。 '[a-zA-Z_\x7f-\xff][a-zA-Z0-9_\x7f-\xff]*'
注意: ここで言うところの文字とはa-z、A-Z、127から255まで (0x7f-0xff)のアスキー文字を意味します。
ティップ: 付録R もご覧になるとよいでしょう。
変数関連の関数に関する情報については、 変数関数リファレンス を参照ください。
$var = 'Bob'; |
PHP 3では、変数は常にその値により代入されていました。 これは、つまり、ある変数にある式を代入する際、元の式の 値全体がコピーされる側の変数にコピーされるということです。 これは、例えば、ある変数の値を他の変数に代入した後で、 これらの変数の1つを変更しても他の変数には影響を与えないという ことを意味します。この種の代入に関するより詳細な情報については、 式 を参照ください。
PHP 4以降では、変数に値の代入を行う別の方法を提供します。それは、 参照による代入 です。 この場合、新規の変数は元の変数を参照するだけです。 (言いかえると、元の変数の"エイリアスを作る"または元の変数を"指す") 新規の変数への代入は、元の変数に影響し、その逆も同様となります。
参照により代入を行うには、代入する変数(ソース変数)の先頭に アンパサンドを加えます。たとえば、次の簡単なコードは 'My name is Bob'を二度出力します。
<?php |
注意すべき重要な点として、名前のある変数のみが参照により代入できる ということがあります。
<?php |
PHP では変数を初期化する必要はありませんが、そのようにするのはとても よいことです。初期化されていない変数の値は、その型のデフォルト値 - FALSE、ゼロ、空の文字列あるいは空の配列となります。
初期化されていない変数のデフォルト値に依存すると、そのファイルを include している別のファイルで同名の変数が使用されていた場合などに 問題を起こします。また、register_globals が on の場合には重大なセキュリティリスク を抱えることになります。初期化されていない変数を使用すると、 E_NOTICE レベルのエラーが発生します。 しかし、初期化されていない配列に要素を追加する場合はエラーにはなりません。 変数が初期化されているかどうかの判断には、isset() を使用します。
add a note
User Contributed Notes
変数
james at custom-made-sites dot com
30-Sep-2010 04:55
justgook at gmail dot com
03-Aug-2010 01:38
I found interstate solution to work with arrays
<?php
$vars['product']['price']=11;
$aa='product';
$bb='price';
echo $vars{$aa}{$bb};
//prints 11
?>
dimitrov dot adrian at gmail dot com
29-Jul-2010 06:52
This is mine type casting lib, that is very useful for me.
<?php
function CAST_TO_INT($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 06:25
[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>"'. $value .'"</td></tr>';
}
}
echo '</table>';
?>
alexandre at nospam dot gaigalas dot net
06-Jul-2007 09:13
Here's a simple solution for retrieving the variable name, based on the lucas (http://www.php.net/manual/en/language.variables.php#49997) solution, but shorter, just two lines =)
<?php
function var_name(&$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
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 $title_default_Dr value="Dr">Dr</option>
</select>
END;
?>
code at slater dot fr
25-Jan-2007 02:10
Here's a pair of functions to encode/decode any string to be a valid php and javascript variable name.
<?php
function label_encode($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);
}
?>
molnaromatic at gmail dot com
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
Mike at ImmortalSoFar dot com
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
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 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
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
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 ]
mike at go dot online dot pt
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
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
?>
ringo78 at xs4all dot nl
14-Jan-2005 12:27
<?php
// 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
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 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 daersys dot net
20-Jan-2004 08:15
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
unleaded at nospam dot unleadedonline dot net
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:
<?php
$error_msg = &$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