试试这个
对 PHP 4 和 PHP 5 的支持
解决方案:
$string ='Bla bla %yada yada% bla bla %yada yada%';
// Count no of %
$count = substr_count($string,'%');
// Valid string pattern
if ( 0 == ($count % 2) ) {
$urlString = $string;
// Iterate for each pair of % to make it link
for ( $i=1; $i <= $count/2 ; $i++ ) {
$urlString = preg_replace('/%/', "<a href='link$i'>", $urlString, 1);
$urlString = preg_replace('/%/', "</a>", $urlString, 1);
}
}
// Invalid string pattern
else {
echo "Invalid string pattern";
}
// Display generated link
echo $urlString;
preg_replace 函数的工作
$str ='Bla bla %yada yada% bla bla %yada yada%';
$newStr = preg_replace('/%/', '', $str, 2);
echo $newStr;
// Output => Bla bla yada yada bla bla %yada yada%
$str ='Bla bla %yada yada% bla bla %yada yada%';
$newStr = preg_replace('/%/', '', strrev($str), 2);
$newStr = strrev($newStr);
echo $newStr;
// Output => Bla bla %yada yada% bla bla yada yada
$str ='Bla bla %yada yada% bla bla %yada yada%';
$newStr = preg_replace('/%/', '', $str);
echo $newStr;
// Output => Bla bla yada yada bla bla yada yada
参考
http://in1.php.net/preg%5Freplace