使用 PHP,我在我的 gmail 和邮件服务器 ID(公司电子邮件 ID)上发送电子邮件附件。我在我的两个电子邮件帐户中都收到了电子邮件,但是在我的公司电子邮件 ID 上,我的电子邮件附件显示错误(它无法正确打开),而在 gmail 中它很好。有什么想法或建议吗??我在这里粘贴我的 php 代码。任何帮助将不胜感激。
HTML 代码:
<form action="contact.php" method="post" name="form1" enctype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="fileAttach" id="file"><br><br>
<input type="submit" name="submit" value="Submit" align="right">
</form>
联系方式.php:
<?php
if($_POST['submit'])
{
$strTo = "mymail@company.com, mymail@gmail.com";
$strSubject = "Attachment file";
$strMessage = "Attachment";
$txtFormEmail = "mymail@gmail.com";
$strSid = md5(uniqid(time()));
$strHeader = "";
$strHeader .= "From: "."<".$txtFormEmail.">\nReply-To: ".$txtFormEmail."";
$strHeader .= "MIME-Version: 1.0\n";
$strHeader .= "Content-Type: multipart/mixed; boundary=\"".$strSid."\"\n\n";
$strHeader .= "This is a multi-part message in MIME format.\n";
$strHeader .= "--".$strSid."\n";
$strHeader .= "Content-type: text/html; charset=utf-8\n";
$strHeader .= "Content-Transfer-Encoding: 7bit\n\n";
$strHeader .= $strMessage."\n\n";
//*** Attachment ***//
if($_FILES["fileAttach"]["name"] != "")
{
$strFilesName = $_FILES["fileAttach"]["name"];
$strContent = chunk_split(base64_encode(file_get_contents($_FILES["fileAttach"]["tmp_name"])));
$strHeader .= "--".$strSid."\n";
$strHeader .= "Content-Type: application/octet-stream; name=\"".$strFilesName."\"\n";
$strHeader .= "Content-Transfer-Encoding: base64\n";
$strHeader .= "Content-Disposition: attachment; filename=\"".$strFilesName."\"\n\n";
$strHeader .= $strContent."\n\n";
}
$flgSend = mail($strTo,$strSubject,$strMessage,$strHeader);
}
?>