Question

这已更新。

在我的网站上，我有一个表格供人们将数据提交到mysql数据库。但我想知道当有人提交新内容时我如何收到电子邮件。

我一直在努力寻找解决方案，但我需要一些帮助。

我有一个不同的联系表单，我使用phpmailer所以我知道它有用。

它会将数据发送到mysql数据库，但它不会向我发送电子邮件。

HTML：

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script type="text/javascript" src="t3.js"></script>

<form action='t2.php' method='post' id="form1">
<div id = "container">  

    <h3>Basic Information</h3><br>

     <div id="main">

    <p><small>Name:</small><input type='text' id="name"       name='name' /><br><br><br>

    <p><small>Email:</small><input type='text' id="email"      name='email'  /><br><br><br>

    <p><small>Comments:</small><textarea  id='comments' name='comments'  rows="5" cols="40"  placeholder="Comments"></textarea> <br><br><br>

    <input type='hidden'  name='action'  value='create' />
   <p><input type="submit" name="submit" id="submit" value="Email Us!" /></p>
    <input type="reset"   name='reset'   value="Reset" class="reset-org"/>
    <ul id="response" />

</div> 
</form>

Javascript t3.js：

$(function() {
// These first three lines of code compensate for Javascript being turned on and off. 
// It simply changes the submit input field from a type of "submit" to a type of "button".

var paraTag = $('input#submit').parent('p');
$(paraTag).children('input').remove();
$(paraTag).append('<input type="button" name="submit" id="submit" value="Email Us Now!" />');

$('#main input#submit').click(function() {
    $('#main').append('<img src="img/ajax/contact/ajax-loader.gif" class="loaderIcon" alt="Loading..." />');

    var name = $('input#name').val();
    var email = $('input#email').val();
    var comments = $('textarea#comments').val();


$.ajax({

        url: 't2.php',
        type: 'POST',
        data: $('#form1').serialize(),
        success: function(results) {
            $('#main img.loaderIcon').fadeOut(1000);
            $('ul#response').html(results);
           }

});
});
})

Php：t2.php

<?php

$name        =$_POST['name'];
$email       =$_POST['email'];
$comments    =$_POST['comments'];

$action = isset($_POST['action']) ? $_POST['action'] : "";

if($action=='create'){ //the the user submitted the form

$query  = "insert into testdb 
            set
                name = '".$mysqli->real_escape_string($_POST['name'])."', 
                email = '".$mysqli->real_escape_string($_POST['email'])."',
                comments = '".$mysqli->real_escape_string($_POST['comments'])."'";


if( $mysqli ->query($query) ) {
    //if saving success
    echo "User was created.";


require 'PHPMailer/PHPMailerAutoload.php'; //replace with path to PHPMailerAutoload.php
$mail = new PHPMailer;



$mail->IsSMTP();                                      // Set mailer to use SMTP
$mail->Host = '****';  // Specify main and backup server
$mail->Port = ***; // set the SMTP port for the GMAIL server
$mail->SMTPAuth = true;                               // Enable SMTP authentication
$mail->Username = '*****';                            // SMTP username
$mail->Password = '*****!';                           // SMTP password
$mail->SMTPSecure = 'tls';                        // Enable encryption, 'ssl' also accepted

$mail->From = $email;
$mail->FromName = $name;
$mail->AddAddress('****,  'Info');  // Add a recipient
$mail->AddReplyTo($email, $name);       



$mail->isHTML(true);                                  // Set email format to HTML

$mail->Subject = 'Here is the subject';
$mail->Body    = 'This is the HTML message body <b>in bold!</b>';

$mail->Send();

}

else{
//if unable to create new record
echo "Database Error: Unable to create record.";
}

//close database connection
$mysqli->close();
}

?>

Answer 1

如果您安装了PHPMailer，则可以执行此操作（编辑以符合您的信息）：

require 'PHPMailerAutoload.php'; //replace with path to PHPMailerAutoload.php

$action = isset($_POST['action']) ? $_POST['action'] : "";

if($action=='create'){

$query  = "insert into organisation 
          set name = '".$mysqli->real_escape_string($_POST['name'])."', 

$data       =$_POST;
$name       =$data['name'];

if( $mysqli ->query($query) ) {
    //if saving success
    echo "User was created.";
    $mail = new PHPMailer;

$mail->isSMTP();                                      // Set mailer to use SMTP
$mail->Host = 'smtp1.example.com;smtp2.example.com';  // Specify main and backup server
$mail->SMTPAuth = true;                               // Enable SMTP authentication
$mail->Username = 'jswan';                            // SMTP username
$mail->Password = 'secret';                           // SMTP password
$mail->SMTPSecure = 'tls';                            // Enable encryption, 'ssl' also accepted

$mail->From = 'from@example.com';
$mail->FromName = 'Mailer';
$mail->addAddress('josh@example.net', 'Josh Adams');  // Add a recipient
$mail->addAddress('ellen@example.com');               // Name is optional
$mail->addReplyTo('info@example.com', 'Information');
$mail->addCC('cc@example.com');
$mail->addBCC('bcc@example.com');

$mail->WordWrap = 50;                                 // Set word wrap to 50 characters
$mail->addAttachment('/var/tmp/file.tar.gz');         // Add attachments
$mail->addAttachment('/tmp/image.jpg', 'new.jpg');    // Optional name
$mail->isHTML(true);                                  // Set email format to HTML

$mail->Subject = $name + ' was Registered';
$mail->Body    = '<b>' + $name + '</b> was registered.  Congrats on the new recruit!';
$mail->AltBody = 'This is the body in plain text for non-HTML mail clients';
}else{
    //if unable to create new record
    echo "Database Error: Unable to create record.";
}

如果您阅读了readme here.

，那会有所帮助

Answer 2

你说当你的脚本收到帖子时你试图发送一封电子邮件，你已经向我们展示了很多代码，它所做的一件事就是尝试发送电子邮件。正如Fred ii所说，你需要学习如何识别和修复的基本语法错误。您在此处向我们展示代码的代码无效。

从PHP发送邮件非常简单：

mail('user@example.com','this is the subject', 'this is the body of the email');

然而，配置工作MTA要困难得多 - 与PHP集成的方式取决于操作系统。这都是well documented in the manual。

当然，如果你想发送比简单的7位ascii文本更复杂的东西，那么使用像swiftmailer或phpmailer这样的工具会减少生活的痛苦 - 但它仍然需要与你的MTA集成工作 - 因此得到mail（）先工作。

表单在提交数据后不会发送电子邮件。 PHPMailer的

2 个答案: