PHP sprintf() no replacing with argument swapping

Question

The following code produces and undefined variable $s instead of "number two"

define("T1","one");
define("T2","two");

$test="number %2$s";

$test=sprintf($test, T1,T2);

echo $test;

John Conde · Accepted Answer · 2013-02-04 18:49:54Z

7

Single quotes solve your problem. Double quotes cause PHP to interpolate your '$' as a variable.

<?php
define("T1","one");
define("T2","two");

$test='number %2$s';

$test=sprintf($test, T1,T2);

echo $test;

answered Feb 4, 2013 at 18:49

John Conde

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2 Comments

Ok so can I not do this in a heredoc statement as that is actually the point of this.

@user1209203: So then why aren't you using a nowdoc instead?