2017-09-14 74 views
0

當我收到一封電子郵件時,我只想將此電子郵件的一部分保存在數據庫中,因此,我使用正則表達式來刪除不需要的部分。但是,這不起作用。PHP preg_replace在沒有硬編碼的情況下無法工作

function emailToMessage($inbox, $emails){ 
$messages = []; 

foreach ($emails as $email_number){ 

    # $overview contient les infos du mail 
    $overview = imap_fetch_overview($inbox,$email_number,0); 
    $headerInfos = imap_headerinfo($inbox,$email_number); 
    $mailHeader = imap_fetchheader($inbox,$email_number); 
    $structure = imap_fetchstructure($inbox,$email_number); 
    $subject = (isset($overview[0]->subject)) ?$overview[0]->subject : 'Pas de sujet'; 
    $sender = $headerInfos->from[0]->mailbox . "@" . $headerInfos->from[0]->host; 
    $body = imap_fetchbody($inbox,$email_number,1); 

    $body = checkSubtype($structure, $body); // decode if the transfert encoding is base64 and change encoding to utf8 
    $text = cleanMail($body, $mailHeader); 

    #conversion de udate en format pouvant etre inséré dans mySql 
    $mailDateAndTime = date("Y-m-d H:i:s", $overview[0]->udate); 

    $message = [ 
     'title'  => $subject, 
     'text'  => $text, 
     'id_sender' => $sender, 
     'mailDateAndTime' => $mailDateAndTime, 
    ]; 

    $messages[] = $message; 
    } 
return $messages; 
} 

function cleanMail($text, $mailHeader){ 
    $clientMail = checkMailClient($mailHeader); 

switch($clientMail){ 
    case "gmail": 
    echo($text . "<br>"); 
    $gmail = preg_replace("/\s+[0-9]{4}-[0-9]{2}-[0-9]{2}\s+[0-9]{2}:[0-9]{2}\s+GMT\+[0-9]{2}:[0-9]{2}\s+(.)*\s+:(?:\s(.)*)*/i", '$1', $text); 
    echo($gmail . "<br>"); 
    return $gmail; 
    break; 

如果

$body = "test 

2017-09-11 11:55 GMT+02:00 XXX : 

> frgthyjuki"; 

回波($文本);和echo($ gmail);既給我:

測試

2017年9月11日11:55 GMT + 02:00 XXX:

> frgthyjuki

但是,如果我申報$文本emailToMessage()(在$ body = checkSubtype($ structure,$ body);和$ text = cleanMail($ body,$ mailHeader);)或cleanMail()這行之間起作用。

switch($clientMail){ 
    $text = "test 

2017-09-11 11:55 GMT+02:00 XXX : 

> frgthyjuki"; 

    case "gmail": 
    echo($text . "<br>"); 
    $gmail = preg_replace("/\s+[0-9]{4}-[0-9]{2}-[0-9]{2}\s+[0-9]{2}:[0-9]{2}\s+GMT\+[0-9]{2}:[0-9]{2}\s+(.)*\s+:(?:\s(.)*)*/i", '$1', $text); 
    echo($gmail . "<br>"); 
    return $gmail; 
    break; 

echo($ text);會給

測試

2017年9月11日11:55 GMT + 02:00 XXX:

> frgthyjuki

和回波($ GMAIL);將給:

測試

爲什麼當變量是硬編碼是隻工作?我能做些什麼來解決它?

UPDATE

function checkSubtype($structure, $body){ 
    $subtype = strtolower($structure->subtype); 

    if($subtype == "alternative") { 
     $encoding = $structure->parts[0]->parameters[0]->value; 
     $encoding = strtolower($encoding); 
     $transferEncoding = $structure->parts[0]->encoding; 

     $body = checkIfBase64($body, $transferEncoding); 
     $body = quoted_printable_decode($body); 
     $body = changeToUTF8($body, $encoding); 
    } else { 
     $transferEncoding = $structure->encoding; 
     $body = checkIfBase64($body, $transferEncoding); 
    } 
    return $body; 
} 

-

function checkIfBase64($text, $transferEncoding){ 
    if($transferEncoding == "3"){ 
     $text = base64_decode($text); 
    } 
    return $text; 
} 

function changeToUTF8($text, $encoding){ 
    if($encoding != "utf-8" && $encoding != 'utf8'){ 
    $text = utf8_encode($text); 
    } 
    return $text; 
} 

UPDATE 2

我雷鳥和橙測試。

我沒有與Thunderbird一起工作(我在網上查了RegEx,這是正確的),但它適用於Orange。

case "orange": 
return preg_replace("/s*>(?:\s(?:.)*)*/i", '$1', $text); 
break; 

我真的不明白...

更新3

我重新嘗試不可怕的解決方案,它與以前的代碼工作。 這很奇怪...

+2

檢查base64事件是否返回您的字符串。 – delboy1978uk

回答

0

這是一個可怕的解決方案,但它的工作原理...

case "gmail": 
    $step1 = preg_replace("/\s+/i", '#', $text); 
    $step2 = preg_replace("/#*[0-9]{4}-[0-9]{2}-[0-9]{2}#+[0-9]{2}:[0-9]{2}#+GMT\+[0-9]{2}:[0-9]{2}#+(.)*#*:(?:#(.)*)*/i", '$2', $step1); 
    return preg_replace("/#/i", ' ', $step2); 
    break; 
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