PHP的sprintf（）沒有與參數交換

Question

下面的代碼生成和不確定的變量$ s，而不是「兩號」代替

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

$test="number %2$s"; 

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

echo $test; 

Answer 1

單引號解決您的問題。雙引號引起PHP插入你的'$'作爲變量。

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

$test='number %2$s'; 

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

echo $test; 

See it working