URL の検証に正規表現を使用しています。この式は JavaScript では非常にうまく機能しますが、PHP ではこのエラーが発生します
A PHP Error was encountered
Severity: Warning
Message: preg_match() [function.preg-match]: Unknown modifier '('
Filename: home/auth.php
Line Number: 1596
A PHP Error was encountered
Severity: Warning
Message: preg_match() [function.preg-match]: Unknown modifier '('
Filename: home/auth.php
Line Number: 1601
これが私の表現です
$pattern ="/^(http|https|ftp)\:\/\/www\.([a-zA-Z0-9\.\-]+(\:[a-zA-Z0-9\.&%\$\-]+)*@)*(\.){1}((25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9])\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[0-9])|([a-zA-Z0-9\-]+\.)*[a-zA-Z0-9\-]+\.(com|edu|gov|int|mil|net|org|biz|arpa|info|name|pro|aero|coop|museum|[a-zA-Z]{2}))(\:[0-9]+)*(/($|[a-zA-Z0-9\.\,\?\'\\\+&%\$#\=~_\-]+))*$/";
これはphp関数です
public function valid_url($data)
{
$data = trim($data);
if(!$data)
{
return TRUE;
}
$pattern ="/^(http|https|ftp)\:\/\/www\.([a-zA-Z0-9\.\-]+(\:[a-zA-Z0-9\.&%\$\-]+)*@)*(\.){1}((25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9])\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[0-9])|([a-zA-Z0-9\-]+\.)*[a-zA-Z0-9\-]+\.(com|edu|gov|int|mil|net|org|biz|arpa|info|name|pro|aero|coop|museum|[a-zA-Z]{2}))(\:[0-9]+)*(/($|[a-zA-Z0-9\.\,\?\'\\\+&%\$#\=~_\-]+))*$/";
$valid = preg_match($pattern,$data);
if(!$valid)
{
$data = "http://".$data;
$valid = preg_match($pattern,$data);
}
if(!$valid)
{
$this->form_validation->set_message('valid_url', 'Please enter a valid URL.');
return FALSE;
}
else
{
return TRUE;
}
}
私は正規表現が苦手なので、問題を理解できませんでした。正規表現の修正を手伝ってください。