ScotlandPHP 2019


Es gibt in PHP zwei Operatoren für Zeichenketten (string). Der erste ist der Verknüpfungsoperator ('.'), dessen Rückgabewert eine zusammengesetzte Zeichenkette aus dem rechten und dem linken Argument ist. Der zweite ist der verknüpfende Zuweisungsoperator ('.='), der das Argument auf der rechten Seite an das Argument der linken Seite anhängt. Siehe Zuweisungs-Operatoren für weitere Informationen.

"Hallo ";
$b $a "Welt!"// $b enthält jetzt den Text "Hallo Welt!"

$a "Hallo ";
$a .= "Welt!";    // $a enthält jetzt den Text "Hallo Welt!"

Siehe auch die Abschnitte über Zeichenketten und String-Funktionen.

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User Contributed Notes 8 notes

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


$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:

$var = 3;

echo "Result: " . ($var + 3);
6 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:


= '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

Stephen Clay
13 years ago
"{$str1}{$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.
hexidecimalgadget at hotmail dot com
9 years ago
If you attempt to add numbers with a concatenation operator, your result will be the result of those numbers as strings.


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

Rafael Serna
3 months 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 (.) :


= [
'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
10 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.
Joseph Alvini
2 months 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");
lci at live dot ca
11 months 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.


$variable1 = "testing";

$variable2 = "We are ".variable1;

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