我一直在一个网站上工作,我想从用户获取输入类型“文件”,并使用php 通过电子邮件将其发送给管理员。 / p>
我的代码如下:
html:
<form class="pure-form" enctype="multipart/form-data" action="resumeUpload.php" method="POST">
<label for="resume_attachment">Upload your resume</label>
<input type="file" class="pure-input-rounded" name="resume_attachment">
<button type="submit" name="resume_send" class="mt-20 mb-20" value="1">Submit</button>
</form>
php:
if($_POST['resume_send'] && isset($_FILES['resume_attachment']))
{
//$from_email = 'sender@abc.com'; //from mail, sender email addrress
$recipient_email = 'xyz@gmail.com'; //recipient email addrress
//Load POST data from HTML form
//$sender_name = $_POST["sender_name"] //sender name
//$reply_to_email = $_POST["sender_email"] //sender email, it will be used in "reply-to" header
//$subject = 'New Resume' //subject for the email
//$message = $_POST["message"] //body of the email
/*Always remember to validate the form fields like this
if(strlen($sender_name)<1)
{
die('Name is too short or empty!');
}
*/
//Get uploaded file data using $_FILES array
$tmp_name = $_FILES['resume_file']['tmp_name']; // get the temporary file name of the file on the server
$name = $_FILES['resume_file']['name']; // get the name of the file
$size = $_FILES['resume_file']['size']; // get size of the file for size validation
$type = $_FILES['resume_file']['type']; // get type of the file
$error = $_FILES['resume_file']['error']; // get the error (if any)
//validate form field for attaching the file
if($file_error > 0)
{
die('Upload error or No files uploaded');
}
//read from the uploaded file & base64_encode content
$handle = fopen($tmp_name, "r"); // set the file handle only for reading the file
$content = fread($handle, $size); // reading the file
fclose($handle); // close upon completion
$encoded_content = chunk_split(base64_encode($content));
$boundary = md5("random"); // define boundary with a md5 hashed value
//header
$headers = "MIME-Version: 1.0\r\n"; // Defining the MIME version
//$headers .= "From:".$from_email."\r\n"; // Sender Email
//$headers .= "Reply-To: ".$reply_to_email."\r\n"; // Email addrress to reach back
$headers .= "Content-Type: multipart/mixed;\r\n"; // Defining Content-Type
$headers .= "boundary = $boundary\r\n"; //Defining the Boundary
//plain text
$body = "--$boundary\r\n";
$body .= "Content-Type: text/plain; charset=ISO-8859-1\r\n";
$body .= "Content-Transfer-Encoding: base64\r\n\r\n";
$body .= chunk_split(base64_encode($message));
//attachment
$body .= "--$boundary\r\n";
$body .="Content-Type: $file_type; name=".$file_name."\r\n";
$body .="Content-Disposition: attachment; filename=".$file_name."\r\n";
$body .="Content-Transfer-Encoding: base64\r\n";
$body .="X-Attachment-Id: ".rand(1000, 99999)."\r\n\r\n";
$body .= $encoded_content; // Attaching the encoded file with email
$sentMailResult = mail($recipient_email, $subject, $body, $headers);
if($sentMailResult )
{
echo "File Sent Successfully.";
unlink($name); // delete the file after attachment sent.
}
else
{
die("Sorry but the email could not be sent.
Please go back and try again!");
}
}
?>
我已使用此代码,但无法完成任务。
任何人都可以在我的代码中找到错误,并提供正确的解决方法。
谢谢。
答案 0 :(得分:2)
您的问题似乎在这一行:
if($_POST['button'] && isset($_FILES['attachment']))
它有两个问题:
首先,$_POST['button']
将始终产生错误的值,因为未设置该值。为此,您需要更改html:<button type="submit" class="mt-20 mb-20" name="button" value="1">Submit</button>
。也就是说:添加一个name
和value
属性,该值应为not be something that results in falsey
。
您打算进行这项检查的目的是什么?如果您只想检查请求是否以POST
的身份进入,请跳过此步骤,因为$_FILES
的存在就足够了。没有$_GET
文件上传。
第二,isset($_FILES['attachment'])
始终为假,因为在HTML中,此文件上载名称为resume_file
,而不是attachment
。
只有解决了这些问题,您才能继续解决邮件问题(如果存在)
要清楚:您的电子邮件代码从未到达。
答案 1 :(得分:-1)
我认为您应该使用PHPMailer API发送带有附件的邮件。
这里有个提示:
要求用户上传具有输入类型文件的文件
将此文件上传到您的服务器
使用PHPMailer发送附件
$mail->addAttachment('/var/tmp/file.tar.gz');