Dizge İşleçleri

İki tane dizge işleci vardır. İlki sol ve sağ terimlerini ard arda ekleyen birleştirme işleci ('.'), ikincisi ise sağ taraftaki bağımsız değişkeni sol taraftaki bağımsız değişkene ekleyen birleştirerek atama işlecidir (.=). Daha fazla bilgi için bkz: atama işleçleri.

<?php
$a
= "Merhaba ";
$b = $a . "Dünyalı!"; // $b artık "Merhaba Dünyalı!" içeriyor

$a = "Merhaba ";
$a .= "Dünyalı!"; // $a artık "Merhaba Dünyalı!" içeriyor
?>

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User Contributed Notes 8 notes

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182
K.Alex
11 years ago
As for me, curly braces serve good substitution for concatenation, and they are quicker to type and code looks cleaner. Remember to use double quotes (" ") as their content is parced by php, because in single quotes (' ') you'll get litaral name of variable provided:

<?php

$a
= '12345';

// This works:
echo "qwe{$a}rty"; // qwe12345rty, using braces
echo "qwe" . $a . "rty"; // qwe12345rty, concatenation used

// Does not work:
echo 'qwe{$a}rty'; // qwe{$a}rty, single quotes are not parsed
echo "qwe$arty"; // qwe, because $a became $arty, which is undefined

?>
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167
anders dot benke at telia dot com
19 years ago
A word of caution - the dot operator has the same precedence as + and -, which can yield unexpected results.

Example:

<php
$var = 3;

echo "Result: " . $var + 3;
?>

The above will print out "3" instead of "Result: 6", since first the string "Result3" is created and this is then added to 3 yielding 3, non-empty non-numeric strings being converted to 0.

To print "Result: 6", use parantheses to alter precedence:

<php
$var = 3;

echo "Result: " . ($var + 3);
?>
up
85
hexidecimalgadget at hotmail dot com
15 years ago
If you attempt to add numbers with a concatenation operator, your result will be the result of those numbers as strings.

<?php

echo "thr"."ee";           //prints the string "three"
echo "twe" . "lve";        //prints the string "twelve"
echo 1 . 2;                //prints the string "12"
echo 1.2;                  //prints the number 1.2
echo 1+2;                  //prints the number 3

?>
up
91
Stephen Clay
18 years ago
<?php
"{$str1}{$str2}{$str3}"; // one concat = fast
 
$str1. $str2. $str3;   // two concats = slow
?>
Use double quotes to concat more than two strings instead of multiple '.' operators.  PHP is forced to re-concatenate with every '.' operator.
up
25
mariusads::at::helpedia.com
15 years ago
Be careful so that you don't type "." instead of ";" at the end of a line.

It took me more than 30 minutes to debug a long script because of something like this:

<?
echo 'a'.
$c = 'x';
echo 'b';
echo 'c';
?>

The output is "axbc", because of the dot on the first line.
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-29
Rafael Serna
5 years ago
Please note that concatenating an array item value using the  key => value syntax will result in a parse error if there are one or more spaces bewteen the concatenation operator (.) :

<?php

$elements
= [
   
'id' => 63,
   
'name' => '{"name": "'. str_repeat("a", 1000) .'"}' // <-- Parse error (there are spaces surrounding string_repeat)
];

$elements = [
   
'id' => 63,
   
'name' => '{"name": "'.str_repeat("a", 1000).'"}' // <-- OK
];
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-46
Joseph Alvini
5 years ago
Concatenation inside of a for/foreach loop. Great for adding onto a string.

function Coffee($string){
    $arr = ["Loves", "Coffee", "With", "His", "Donuts"];
 
    foreach($arr as $items){
          $string  .= ' ' . $items;
    }
   
    return $string;
}

echo Coffee("Joe");
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-64
lci at live dot ca
6 years ago
Note that the . operator accepts unquoted strings/undefined identifiers.

So $var = "test".test; will result in "testtest" being written to $var and not an error (as one might expect).

Be careful when trying to concatenate variables with strings, if you miss the $ before the variable name it will just concatenate the string with the variable name.

Example:

$variable1 = "testing";

$variable2 = "We are ".variable1;

$variable2 is now "We are variable1" as opposed to the intended "We are testing".
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