有两个字符串（string）运算符。第一个是连接运算符（“.”），它返回其左右参数连接后的字符串。第二个是连接赋值运算符（“.=”），它将右边参数附加到左边的参数之后。更多信息见赋值运算符。
<?php
$a = "Hello ";
$b = $a . "World!"; // now $b contains "Hello World!"
$a = "Hello ";
$a .= "World!"; // now $a contains "Hello World!"
?>
K.Alex (2012-12-26 22:22:03)
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
?>
hexidecimalgadget at hotmail dot com (2009-02-09 21:37:33)
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
?>
mariusads::at::helpedia.com (2008-08-27 02:44:46)
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'.
$c = 'x';
echo 'b';
echo 'c';
?>
The output is "axbc", because of the dot on the first line.
Stephen Clay (2005-12-23 07:10:56)
<?php
"{$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.
anders dot benke at telia dot com (2004-04-27 09:53:35)
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);
?>