Question

我终于制作了我的脚本，获取图像的链接并在主要照片中显示。这是链接 http://www.fisicasanitaria.it/form/form.php 当表单在另一页面中提交页面重定向时 http://www.fisicasanitaria.it/form/welcome.php

这是此页面的代码

<link rel="stylesheet" href="form.css">

<script type="text/javascript" src="js/jquery-3.1.0.min.js"></script>
<script src="js/script.js"></script>

<?php 
/* welcome.php */

//$_SESSION variables become available on this page
session_start(); 
?>
<div class="body content">
<div class="welcome">
<div class="intro">


<div class="alert alert-success"><?= $_SESSION['message'] ?></div>
    <img class="dim" src="<?= $_SESSION['avatar'] ?>" ><br />
    Welcome <span class="user"><?= $_SESSION['username'] ?></span>
    Welcome <span class="email"><?= $_SESSION['email'] ?></span>
    Welcome <span class="password"><?= $_SESSION['password'] ?></span>
    </div>
    <section id="content"></section>

    <?php
    $mysqli = new mysqli("myHost", "myUser", "myRoot", "database");
    $sql = "SELECT username, avatar,email,password, id FROM users";
    //$result = mysqli_result object
    $result = $mysqli->query( $sql ); 
    ?>
    <div id='registered'>

    <span>All registered users:</span>
    <?php
    //returns associative array of fetched row
    while( $row = $result->fetch_assoc() ){ 
        echo "<div class='userlist'><span>".$row['username']."</span><span>".$row['email']."</span><br />";
        echo "<a class='prova'   >{$row['user']}<img class='foto' id=\"{$row['id']}\" src='".$row['avatar']." '></a></div>";
      // echo "<a href=\"welcome.php?id={$row['id']}\">{$row['user']}</a>";
    }
?>  
</div>
</div>
</div>

使用此脚本我得到底部图像的链接，我想显示并替换主要的照片。

jQuery(function () {
    jQuery(" #registered img").hover(function(e) {
        var currentId = jQuery(this).attr("src");
        var ind = currentId;
        jQuery("body > div > div > div.intro > img").fadeOut();
        jQuery("body > div > div > div.intro > img").attr("src",ind).fadeIn();
        e.preventDefault();   
    });
});

为什么foto会出现两次这种奇怪的效果？我不太了解命令preventDefault，这是正确的吗？谢谢

奇怪的效果出现两次

0 个答案: