Ajax联系人表单不能正常工作


Ajax contact form not working properly

我使用的是ajax联系人表单,但是它似乎不能正常工作。当我尝试提交表单时,我收到"请填写所有字段"的消息,尽管我已经填写了所有字段。

下面是我当前的表单代码:
    <form method="get" action="mail.php">
    <input name="fname" type="text" value="Enter your First Name here..." id="fname" onfocus="if(this.value == 'Enter your First Name here...') this.value = ''"/>
    <input name="lname" type="text" value="Enter your Last Name here..." id="lname" onfocus="if(this.value == 'Enter your Last Name here...') this.value = ''"/>
    <input name="cname" type="text" value="Enter your Company Name here..." id="cname" onfocus="if(this.value == 'Enter your Company Name here...') this.value = ''"/>
    <input name="web" type="text" value="Enter your Website here..." id="web" onfocus="if(this.value == 'Enter your Website here...') this.value = ''"/>
    <input name="email" type="text" value="Enter your E-mail here..." id="email" onfocus="if(this.value == 'Enter your E-mail here...') this.value = ''"/>
    <input name="phone" type="text" value="Enter your Phone Number here..." id="phone" onfocus="if(this.value == 'Enter your Phone Number here...') this.value = ''"/>
    <textarea name="message" id="message" onfocus="if(this.value == 'Enter your message here...') this.value = ''" rows="" cols="">Enter your message here...</textarea>
    <input name="submit" type="submit" value="Submit" id="submit"/>
    <div id="response" style="float:right">
    <p></p>
    </div>
    </form> 

my mail.php:

    <?php
    if ($_GET['fname'] == "" || $_GET['lname'] == "" || $_GET['cname'] == "" || $_GET['web'] == "" || $_GET['phone'] == "" ||  $_GET['email'] == "" || $_GET['message'] == "") {
        echo 'Please fill in all fields!';
        exit;
    }
    if (strlen($_GET['fname']) < 3) {
        echo 'Your name should be atleast 3 characters long!';
        exit;
    }
    if (strlen($_GET['lname']) < 3) {
        echo 'Your name should be atleast 3 characters long!';
        exit;
    }
    if (strlen($_GET['cname']) < 3) {
        echo 'Your name should be atleast 3 characters long!';
        exit;
    }
    if (strlen($_GET['web']) < 3) {
        echo 'Your name should be atleast 3 characters long!';
        exit;
    }
    if (strlen($_GET['phone']) < 3) {
        echo 'Your name should be atleast 3 characters long!';
        exit;
    }
    if (!preg_match("/^([a-zA-Z0-9])+(['.a-zA-Z0-9_-])*@([a-zA-Z0-9_-])+('.[a-zA-Z0-9_-]+)*'.([a-zA-Z]{2,6})$/", $_GET['email'])) {
        echo 'Please enter a valid email address!'; 
        exit;
    }
    if (strlen($_GET['message']) < 10) {
        echo 'Your message should be atleast 10 characters long!';
        exit;
    }
    $to = 'my@email.com';
    $name = stripslashes($_GET['name']);
    $email = stripslashes($_GET['email']);
    $subject  = "Contact request";
    $msg  = "From:".$name."'r'n";
    $msg .= "e-Mail:".$email."'r'n";
    $msg .= "Subject:".$subject."'r'n'n";
    $msg .= "---Message--- 'r'n".nl2br(stripslashes($_GET['message']))."'r'n'n";
    $mail = mail($to, $subject, $msg, "From:".$email);
    if($mail) {
        echo 'We will be in touch shortly!';
    } else {
        echo 'Error: Message could not be sent!';
    }
    ?>

最后但并非最不重要的ajax.js

     $(function() {
        $("input#submit").click( 
            function() {
                $('#response p').empty().append("<img src='img/ajax-loader.gif' />");
                var fname = $("#fname").val();  
                var lname = $("#lname").val();
                var cname = $("#cname").val(); 
                var web = $("#web").val();  
                var email = $("#email").val(); 
                var phone = $("#phone").val(); 
                var message = $("#message").val(); 
                var dataString = 'fname='+ fname + 'lname='+ lname + 'cname='+ cname + 'web='+ web + '&email=' + email + 'phone='+ phone + '&message=' + message;  
                $.ajax({
                    url: "mail.php",
                    data: dataString,
                    type: "GET",
                    success: function(responseText) {
                         $('#response p').empty().append(responseText);
                         }
                     });
            return false;
        });
      });

你知道这是怎么回事吗?一些专家的建议会很有帮助。

在构建dataString时,您可能需要在每个参数前加上&

var dataString = 'fname='+ fname + '&lname='+ lname + '&cname='+ cname + '&web='+ web + '&email=' + email + '&phone='+ phone + '&message=' + message;

在您的代码中,您只对emailmessage这样做。