與stackoverflow開發人員,我贏了我的第一個任務。我正在創建傳入的電子郵件插入到數據庫查詢。當傳入的郵件插入到數據庫時,只有一個電子郵件詳細信息插入到數據庫
這是我的代碼:
#!/usr/bin/php -q
<?PHP
/* connect to gmail */
$hostname = '{example.org:995/pop3/ssl/novalidate-cert}';
$username = 'user';
$password = 'pass';
/* try to connect */
$inbox = imap_open($hostname,$username,$password) or die('Cannot connect to SERVER: ' . imap_last_error());
/* grab emails */
$emails = imap_search($inbox,'ALL');
/* if emails are returned, cycle through each... */
if($emails) {
/* begin output var */
$output = '';
/* put the newest emails on top */
rsort($emails);
/* for every email... */
foreach($emails as $email_number) {
/* get information specific to this email */
$overview = imap_fetch_overview($inbox,$email_number,0);
$message = imap_fetchbody($inbox,$email_number,2);
/* output the email header information */
//$read_status = '<div class="toggler '.($overview[0]->seen ? 'read' : 'unread').'">';
$subject = $overview[0]->subject;
$from = $overview[0]->from;
$received_date = '<span class="date">on '.$overview[0]->date.'</span>';
//$output.= '</div>';
/* output the email body */
$output= '<div class="body">'.$message.'</div>';
}
//echo $output;
$servername = "localhost";
$username = "username ";
$password = "password ";
$dbname = "dbname";
$conn = mysqli_connect($servername, $username, $password, $dbname) or die("Connection failed: " . mysqli_connect_error());
//INSERT INTO DATABASE
$sql = "INSERT INTO ads(`banner_name`, `hd_image`)VALUES('".$message."', '".$subject."') ";
$result = mysqli_query($conn, $sql);
}
/* close the connection */
imap_close($inbox);
?>
這是運作良好。我的問題是,當我發送一封新郵件到我的電子郵件地址。查詢將每次執行並插入第一封電子郵件詳細信息。不新發送電子郵件的詳細信請幫我解決這個問題。
另外,如果電子郵件收到附件,我該如何下載?
感謝您的幫助很大
您將從循環中獲得最終的項目(如果有多個電子郵件,您的代碼只會將最後一個插入到數據庫中)。在foreach循環中執行數據庫操作。 – Nikhil
嗨, 我把DB代碼從循環中,但仍然收到我的第一封電子郵件,而不是接收新發送的詳細信息 謝謝! –
*在foreach循環中* –