所以我試圖讓這個小腳本工作的樂趣,只是不能得到它的工作, 我有一個隨機名稱分配他們的'隨機'電子郵件地址數組,我想檢查電子郵件地址是否有擴展名: '@ hotmail.com', '@ hotmail.ca',或 '@ yahoo.ca'。如果他們接着迴應類似'丹尼爾,你的電子郵件地址擴展名是$擴展名,你有資格' ,如果他們不只是說'凱拉,你的電子郵件地址是$擴展名,你沒有資格。返回數組中的字符串的一部分
我想添加一個foreach語句來解釋數組$ classRoom2中的每個人,我嘗試過使用strstr(),但它不能在foreach內工作,因爲它只能有一個字符串。
繼承人什麼我走到這一步:
qualify = "";
$ClassRoom2 = array(
'Daniel' => '[email protected]',
'Mike' => 'dkoko[email protected]',
'Meranda' => '[email protected]',
'Will' => '[email protected]',
'Brittey' => '[email protected]',
'Kayla' => '[email protected]');
switch ($ClassRoom2) {
case '@hotmail.com':
echo 'You are using extension '. $final; $qualify = 1; break;
case '@hotmail.ca':
echo 'You are using extension '. $final; $qualify = 1; break;
case '@yahoo.com':
echo 'You are using extension '. $final; $quality = 1; break;
case '@yahoo.ca':
echo 'You are using extension '. $final; $qualify = 1; break;
case '@live.ca':
echo 'You are using extension '. $final; $quality = 1; break;
default:
echo 'its something else'; $qualify = 0;
break;
}
if ($qualify == 1) {
echo "Congratulations, you quality for the contest. The extension you chose was <b>$final</b>";
} else {
echo "Sorry mate! you didn't quality for the contest.";
}
使用'爆炸( '@', '[email protected]')'.. – check
'strrpos('[email protected]」 ,'@ hotmail.com',0)=== 0' –
您可能想要使用[loop](http://php.net/manual/en/language.control-structures.php)和[regular表達](http://php.net/manual/en/book.pcre.php),例如'@hotmail \ [A-Z] {2,3}'。看看['foreach()'](http://php.net/manual/en/control-structures.foreach.php)和['preg_match()'](http://php.net/manual /en/function.preg-match.php) – Havelock