Question

好的，我不能让这个jQuery无论如何都能运行！它就这么简单！我想要它做的就是让#main div在加载时淡入淡出。

<!DOCTYPE html>
<html>
<head>

<script type="text/javascript" src="randomizer.js"></script>

<title>
<?php
if (isset($_POST['first']) && isset($_POST['second'])) {
     echo "Randomized {$_POST['first']}-{$_POST['second']}";
}
else {
     echo "Randomizer";
}
?>
</title>

<link type="text/css" rel="stylesheet" href="randomizer.css" />
<script type="text/javascript" src="randomizer.js"></script>
</head>
<body>

<?php
     if (isset($_POST['first']) && isset($_POST['second'])) {
         $grabFirst = strip_tags($_POST['first']);
         $grabSecond = strip_tags($_POST['second']);
     }
?>
<div id="main">
<center><form action="randomizer.php" method="post">
     <input type="text" name="first" value="<?php if(isset($grabFirst)){ echo htmlspecialchars($grabFirst); } ?>" placeholder="First Number"  /><br />
     <input type="text" name="second" value="<?php if(isset($grabSecond)){ echo htmlspecialchars($grabSecond); } ?>" placeholder="Second Number" /><br />
     <input type="submit" />
</form>

<?php

     if (isset($_POST['first']) && isset($_POST['second'])) {
         $grabFirst = strip_tags($_POST['first']);
         $grabSecond = strip_tags($_POST['second']);
     }
     else {
         print "Please enter two numbers.\n";
         die;
     }

     if (strlen($grabFirst) < 1 or strlen($grabSecond) < 1) {
         print "Please make sure you filled out both feilds.\n";
         die;
     }
     elseif (!is_numeric($grabFirst) && !is_numeric($grabSecond)) {
         print "Please make sure you entered two valid numbers.\n";
         die;
     }
     elseif (!is_numeric($grabSecond)) {
         print "Please make sure you entered two valid numbers.\n";
         die;
     }
     elseif (!is_numeric($grabFirst)) {
         print "Please make sure you entered two valid numbers.\n";
         die;
     }
     else {
     }

     $rand = rand($grabFirst, $grabSecond);

     print "A random number between {$grabFirst} and {$grabSecond} is {$rand}.\n";
?></center>
</div>
</body>
</html>

继承我的jQuery

$(document).ready(function(){
     $('#main').fadeTo('slow', 1);
});

如果有人能提供帮助，我真的很感激。我一直想把它修好一个小时，我没有运气。我刚开始使用jQuery而没有经验，也许这是一个非常愚蠢的错误，但我不知道。感谢〜

Answer 1

您需要在页面中包含jQuery：

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>

为什么jQuery不工作？

1 个答案: