我如何在PHP中使用Jquery

时间:2015-10-20 09:48:58

标签: php jquery

我需要为学校创建一个数据库。当我在我的专栏“Champion”中悬停一个物体时,我想向用户显示左侧冠军的照片。我想用Jquery做这个。我在网上搜索过,但我找不到答案。这是我的剧本:

Observable.create({
    ->
    // doing background task...
})
.onSubscribe(Schedules.newThread())
.flatMap(() -> getOtherObservable())
.subscribe(() -> {
    // onNext...
});

private Observable getOtherObservable(){
    // return other Observable
    return ...;
}

1 个答案:

答案 0 :(得分:0)

首先,我看到你有点困惑。

PHP 是一种server-side编程语言;这意味着在将页面发送到客户端(浏览器)之前,所有PHP代码都在服务器上呈现。

JQuery 是一个JavaScript类,旨在简化JavaScript脚本/编程; JavaScript是一种client-side脚本/编程语言,这意味着任何JavaScript/JQuery代码都在客户端上执行(我们可以说是浏览器)。

好的,现在关于你的问题。如果要在一组单元格/列的鼠标悬停事件中显示图像,可以执行以下操作:

<html>
<head>
    <link rel="stylesheet" type="text/css" href="stijl.css">
</head>
<body>    
        <!-- the following code does not need PHP, so just leave it as HTML -->
        <aside id="links">
            <p id="Aatrox">
                <img src="http://eindwerkstef.be/php/champions/Aatrox.jpg"/>
            <p>
        </aside>
            <form action="php54.php" method="post">
                <aside>
                    <select name="sorteren">
                        <option value="Champion">Champion</option>
                        <option value="Role">Role</option>
                        <option value="Lane">Lane</option>
                        <option value="Counter">Counter</option>
                    </select>
                    <select name="sorteren2">
                        <option value="Champion">Champion</option>
                        <option value="Role">Role</option>
                        <option value="Lane">Lane</option>
                        <option value="Counter">Counter</option>
                    </select>
                    </br>
                    <input type="submit" name="submit" value="Sorteer!"/>
                    </br>
                    <p>Wat wilt u Weergeven?</p>
                    <select name="toonL">
                        <option value="Top">Top</option>
                        <option value="Mid">Mid</option>
                        <option value="Jungle">Jungle</option>
                        <option value="Bottom">Bottom</option>
                    </select>
                    <input type="submit" name="toon1" value="Toon"/>
                </aside>
    <?php
        $address = "localhost"; // this is your database's address
        $username = "root"; // this is your mysql username
        $password = "1234"; // this is your mysql password
        $database = "example"; // this is your database
        $connection = mysqli_connect($address, $username, $password, $database);

        if (isset($_POST['submit'])){
            $query = "SELECT * FROM lolOverzicht ORDER BY {$_POST['sorteren']}, {$_POST['sorteren2']}";
            $res = mysqli_query($connection, $query);
            echo("<table border='1' id='tbloverzicht'><tr><td><b>Champion</td><td><b>Role</td><td><b>Lane</td><td><b>Counter</td></tr>");

            while ($row = $res->fetch_assoc()){
                echo("<tr>");
                echo("<td class='col-champion'>".$row['Champion']."</td>
                      <td>".$row['Role']."</td>
                      <td>".$row['Lane']."</td>
                      <td>".$row['Counter']."</td>
                      <td><a href=\"lolwijzig.php?id={$row['Nummer']}\">Wijzig</a></td>
                      <td><a href=\"lolverwijder.php?id={$row['Nummer']}\">Verwijder</a>
                      </td>");
                echo("</tr>");
            }
            echo "</table>";
        }

        if (isset($_POST['toon1'])){
            $query = "SELECT * FROM lolOverzicht WHERE Lane = '{$_POST['toonL']}' ORDER BY {$_POST['sorteren']}, {$_POST['sorteren2']}";
            $res = mysqli_query($connection, $query);
            echo "<table border='1' id='tbloverzicht'><tr><td><b>Champion</td><td><b>Role</td><td><b>Lane</td><td><b>Counter</td></tr>";

            while ($row=mysql_fetch_array($res)){
                echo("<tr>");
                echo("<td class='col-champion'>".$row['Champion']."</td>
                      <td>".$row['Role']."</td>
                      <td>".$row['Lane']."</td>
                      <td>".$row['Counter']."</td>
                      <td><a href=\"lolwijzig.php?id={$row['Champion']}\">Wijzig</a></td>
                      <td><a href=\"lolverwijder.php?id={$row['Champion']}\">Verwijder</a></td>");
                echo("</tr>");
            }
            echo("</table>");
        }        
        mysqli_close($connection);
    ?>

    <img src"" id="img">

    <script src="script/jquery-1.11.1.min.js" type="text/javascript"></script>
    <script>
        $(document).ready(function(){
            $('#img').hide();
        });

        $('.col-champion').mouseover( function(e){
            /* 
            Since I don't know where and how you decide which image to show, that's up to you 
            then to change img source just use:
            $("#img").attr("src","chosenimage.jpg");
            */
            $('#img').show();
            $("#img").offset({left:e.pageX,top:e.pageY});
        });

        $('.col-champion').mouseout( function(e){
            $('#img').hide();
        });

    </script>    
</body>

注意:我从mysql更改为mysqli,因为现在不推荐使用mysql,我强烈建议您使用mysqli进行下一个项目。

建议:对于未来的问题,我强烈建议您解释问题的每一个细节,提供您认为可以使您的问题更容易理解的所有内容。

然后,如果您需要一些淡入淡出效果,可以点击here

查看JQuery的官方文档。

我希望这能帮到你,祝你的功课好运!