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让我提前道歉,我就像昨天一样跳进了jquery的事情,所以慢慢说,使用很多初学者术语。无论如何,我试图让我的 jquery/ajax 联系表单停止刷新并保持在同一页面上,但我一直收到错误消息“ReferenceError: $ is not defined”。我试图解决这个问题并弄清楚它是否与我在这里的失败有关。

这是HTML……

<div id="contact">
            <h2>Contact</h2>
            <div class="clear"></div>
            <div class="contactContainer">
                <h3>Get in touch:</h3>
                <div class="contact">

                    <form action="js/ajaxcontactform/sendmail.php" method="post" id="contactForm">
                    <ul>
                    <li>
                    <label for="name">Name:<font color="#ff3300">*</font></label>
                    <input type="text" name="name" value="" id="name" />
                    </li>
                    <li>
                    <label for="email">Email:<font color="#ff3300">*</font></label>
                    <input type="text" name="email" value="" id="email" />
                    </li>
                    <li>
                    <label for="tele">Telephone:</label>
                    <input type="text" name="tele" value="" id="tele" />
                    </li>
                    <li class="special" style="display: none;">
                    <label for="last">Don't fill this in:</label>
                    <input type="text" name="last" value="" id="last" />
                    </li>
                    <li>
                    <label for="message">Message:<font color="#ff3300">*</font></label><textarea rows="5" name="message"></textarea>
                    </li>
                    <li class="submitbutton">
                    <input type="submit" class="btn" value="Send Message" />
                    </li>
                    </ul>
                    </form>

这里是php.....

    <?php

// basic settings section
$sendto = 'your@email.com';
$subject = 'You have a new message from your virtual resume!';
$iserrormessage = 'There was a problem with sending e-mail to us, please check:';
$thanks = "Thank's for your message! I'll contact you as soon as possible!";

$emptyname = 'Did you enter your name?';
$emptyemail = 'Did you enter your e-mail address?';
$emptymessage = 'Did you enter the message?';
$emptyphone = 'Did you enter phone number?';

$alertname = 'Please enter your name with standard alphabet!';
$alertemail = 'Please enter your e-maill address in format: name@domain.com';
$alertmessage = "Please do not use any parenthesis or other escaping characters. Standard web url's should work fine!";
$alertphone = 'Please enter your phone number without any special characters, only numbers ex: 5553212';

$alert = '';
$iserror = 0;

// cleaning the post variables
function clean_var($variable) {$variable = strip_tags(stripslashes(trim(rtrim($variable))));return $variable;}

// validation of filled form
if ( empty($_REQUEST['name']) ) {
$iserror = 1;
$alert .= "<li>" . $emptyname . "</li>";
} elseif ( ereg( "[][{}()*+?.\\^$|]", $_REQUEST['name'] ) ) {
$iserror = 1;
$alert .= "<li>" . $alertname . "</li>";
}


if ( empty($_REQUEST['email']) ) {
$iserror = 1;
$alert .= "<li>" . $emptyemail . "</li>";
} elseif ( !eregi("^[_a-z0-9-]+(.[_a-z0-9-]+)*@[a-z0-9-]+(.[a-z0-9-]+)*(.[a-z]{2,3})$", $_REQUEST['email']) ) {
$iserror = 1;
$alert .= "<li>" . $alertemail . "</li>";
}

if ( empty($_REQUEST['message']) ) {
$iserror = 1;
$alert .= "<li>" . $emptymessage . "</li>";
} elseif ( ereg( "[][{}*+\\^|]", $_REQUEST['message'] ) ) {
$iserror = 1;
$alert .= "<li>" . $alertmessage . "</li>";
}

// if there was error, print alert message
if ( $iserror==1 ) {

echo "<script type='text/javascript'>$(\".message\").hide(\"slow\").fadeIn(\"slow\").delay(5000).fadeOut(\"slow\"); </script>";
echo "<strong>" . $iserrormessage . "</strong>";
echo "<ul>";
echo $alert;
echo "</ul>";

} else {
// if everything went fine, send e-mail

$msg = "From: " . clean_var($_REQUEST['name']) . "\n";
$msg .= "Email: " . clean_var($_REQUEST['email']) . "\n";
$msg .= "Message: \n" . clean_var($_REQUEST['message']);
$header = 'From:'. clean_var($_REQUEST['email']);

mail($sendto, $subject, $msg, $header);

echo "<script type='text/javascript'>$(\".message\").fadeOut(\"slow\").fadeIn(\"slow\").animate({opacity: 1.0}, 5000).fadeOut(\"slow\");</script>";
echo $thanks;

die();
}
?>

我使用标准的 jquery.form.js(据我所知)

再次提前感谢!

4

2 回答 2

1

你导入 jquery 库吗?试着把它放在你的代码头中。

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
于 2013-10-15T19:21:01.030 回答
1

查看您的实际站点,我发现并非所有 JS 包含在您指定的文件夹中。特别是对于jquery.form.js. 由于没有找到这个插件,所以提交表单时页面自然会刷新。

查看 HTML,我发现了这一行:

<script type="text/javascript" src="/js/ajaxcontactform/jquery.form.js"></script>

如果您查找http://www.robergeaz.com/js/ajaxcontactform/jquery.form.js,您会发现它不存在。

看看你的其他 JS 位置,我发现它们中的大多数都使用相对路径js/xyz.js,这相当于http://www.robergeaz.com/darin/js/xyz.js. 使用与表单插件的其他插件相同的相对路径会在http://www.robergeaz.com/darin/js/ajaxcontactform/jquery.form.js.

因此,要解决您的问题,请将上述脚本更改为:

<script type="text/javascript" src="js/ajaxcontactform/jquery.form.js"></script>

/注意开头的缺失src

编辑:

此外,您需要在包含主 jQuery 库之后加载插件。

于 2013-10-15T20:02:57.530 回答