Question

我正在尝试使用php表单和AJAX发送简单的电子邮件。但是，每次执行此操作时，代码都会抛出email could not be sent, please try again错误。为什么会这样？

我还尝试alert警报序列中的变量，但它不起作用。我检查了变量并且它们似乎有效，但似乎我忽略了一些东西。

Javascript / jQuery （在<head>区块中）

<script type="text/javascript" src="js2/jquery-1.7.2.min.js"></script> 
<script>
$("#form2").submit(function(e){
    $(".error").remove();    
    $('#cancel2').click(function(){          
        return;
    });    

    var hasError = false;
    var emailReg = /^([A-Za-z0-9_\-\.])+\@([A-Za-z0-9_\-\.])+\.([A-Za-z]{2,4})$/;

    var nameVal = $("#name").val();

    if (nameVal == '') {
        $("#name").after('<span class="error">You forgot to enter the name.</span>');
        hasError = true;
    }

    var emailToVal = $("#emailTo").val();

    if (emailToVal == '') {
        $("#emailTo").after('<span class="error">You forgot to enter the email address to send to.</span>');
        hasError = true;
    } else if(!emailReg.test(emailToVal)) {
        $("#emailTo").after('<span class="error">Enter a valid email address to send to.</span>');
        hasError = true;
    }

    var emailFromVal = $("#emailFrom").val();

    if(emailFromVal == '') {
        $("#emailFrom").after('<span class="error">You forgot to enter the email address to send from.</span>');
        hasError = true;
    } else if (!emailReg.test(emailFromVal)) {
        $("#emailFrom").after('<span class="error">Enter a valid email address to send from.</span>');
        hasError = true;
    }

    if (hasError == false){
        $('.navLinks').hide();
        $('.navLinks:before').hide();
        $('.navLinks:after').hide();                 
        s                       
        $.ajax({
            "type":"POST",
            "url":"sendemail.php",
            "data": { name: nameVal, emailTo: emailToVal, emailFrom: emailFromVal },
            "dataType":'json',
            "success":function(){
                //$('#sumbit1').after('<img src="/img/wait.gif" alt="wait" />');    
                alert('Your email has been sent! For faster answer please call us at: (403) 454-5526');
                $('.navLinks').hide();    
             },          
             "error": function (){
                 alert('Your email could not be sent, please try again.');
             }    
        });
    }          

    e.preventDefault();
    return;
});
</script>

HTML

<form name="send-to-friend" id="form2" class="form" method="post" action="sendemail.php">
    <label for="name" class="blockItem">Your Name</label>
    <input type="text" id="name" class="blockItem" name="your name" maxlength="60" size="35"/>
    <label for="emailFrom" class="blockItem">Your Email Address</label>
    <input type="text" id="emailFrom" class="blockItem" name="your email" maxlength="60" size="35" />
    <label for="emailTo" class="blockItem">Your Friend's E-mail Address</label>
    <input  type="text" id="emailTo" class="blockItem" name="your friends email" maxlength="60" size="35"  />
    <input class="button" id="submitt2" type="submit" value="SUBMIT" />                               
</form>

PHP

<?php

$mailTo = $_POST['emailToVal'];
$mailFrom = $_POST['emailFromVal'];
$name = $_POST['nameVal'];
$message = 'I found this company recommended for residential and office cleaning in Calgary. Their website is FineMaid "dot" com (replace "dot", of course in browser\s address bar). Cheers!';  

$subject = $name . ' recommends Fine Maid';

$send = @mail($mailTo, $subject, $message, "From: ".$mailFrom);    

?>

Answer 1

您在PHP中访问的参数不存在，应该是：

$mailTo = $_POST['emailTo'];
$mailFrom = $_POST['emailFrom'];
$name = $_POST['name'];

但是，当您使用JSON作为dataType时，应返回JSON。

此外：当PHP脚本返回有效的JSON时，没有检查邮件是否已经发送，当前jQuery将报告成功（它没有，因为响应至少是一个空格）或者缺少POST变量的注意事项，而不是JSON）

将它放在脚本的末尾：

if(!$send)
{
  header("HTTP/1.0 500 Internal Server Error");
}
else
{
 die(json_encode(array('msg'=>'Your email has been sent! 
                               For faster answer please call us at: (403) 454-5526')));
}

...当$ send为true（邮件已被发送）时，脚本将返回JSON，否则响应将具有HTTP状态代码500，这将触发错误回调

无法使用Ajax / PHP发送电子邮件

1 个答案: