K.Alex ¶5 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
?>
anders dot benke at telia dot com ¶13 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);
?>
Stephen Clay ¶12 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.
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.
<?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 ¶9 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.
lci at live dot ca ¶2 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.
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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