Question

我有一个页面，它从ajax响应加载它的内容。问题是ajax从我的html页面中的head标签中igonres我的所有脚本。有没有办法让这些javascript文件在我的ajax响应中执行？

AJAX：

$(document).ready(function () {
    $(".all").click(function () {
        var all = $(this).attr("id");
        if (all != '') {
            $.ajax({
                type: "POST",
                url: "php/searchbrowselist.php",
                data: "all=" + all,
                async: true,
                success: function (option) {
                    var $this = $("#browsemusic")
                    $this.html(option);
                    $('#sortable1, #sortable2').sortable({
                        connectWith: ".connected"
                    }).disableSelection();
                }
            });
        }
    });
});

PHP：

<?php
include ('dbcon.php');

if (isset($_REQUEST['all']) && $_REQUEST['all'] != '') {

    // ===============================Button "ALL"====================================

    unset($_REQUEST['kw']);
    unset($_REQUEST['genre']);
    $query = "select * from music";
    $result = mysqli_query($link, $query) or die(mysqli_error());
    echo '<ul id="sortable1" class="connected">';
    while ($info = mysqli_fetch_array($result)) {
        echo '<li><div class="ui360"><button type="button" class="addtoplaylist" >Add</button><a href="' . $info['path'] . '"> ' . $info['artist'] . ' - ' . $info['title'] . ' (' . $info['album'] . ') ' . '</a></div><hr /></li>';
    };
    echo '</ul>';
}
elseif (isset($_REQUEST['kw']) && $_REQUEST['kw'] != '') {

    // ============================= Search for music ================================

    $kws = $_REQUEST['kw'];
    $kws = mysqli_real_escape_string($link, $kws);
    $query = "select * from music where title like '%" . $kws . "%' or artist like '%" . $kws . "%'";
    $result = mysqli_query($link, $query) or die(mysqli_error($link));
    echo '<ul id="sortable1" class="connected">';
    while ($info = mysqli_fetch_array($result)) {
        echo '<li><div class="ui360"><button type="button" class="addtoplaylist" >Add</button><a href="' . $info['path'] . '"> ' . $info['artist'] . ' - ' . $info['title'] . ' (' . $info['album'] . ') ' . '</a></div><hr /></li>';
    };
    echo '</ul>';
}
elseif (isset($_REQUEST['genre']) && $_REQUEST['genre'] != '') {

    // =====================================Browse By Genre ===========================================

    $genre = $_REQUEST['genre'];
    $genre = mysqli_real_escape_string($link, $genre);
    $gquery = "select music_id from musicgenre where genre_id = '$genre'";
    $results = mysqli_query($link, $gquery) or die(mysqli_error($link));
    $music = array();
    while ($id_result = mysqli_fetch_array($results)) {
        $music[] = $id_result['music_id'];
    };
    echo '<ul id="sortable1" class="connected">';
    foreach($music as $song) {
        $query = "select * from music where music_id = '$song'";
        $result = mysqli_query($link, $query) or die(mysqli_error());;
        while ($info = mysqli_fetch_array($result)) {
            echo '<li><div class="ui360"><button type="button" class="addtoplaylist" >Add</button><a href="' . $info['path'] . '"> ' . $info['artist'] . ' - ' . $info['title'] . ' (' . $info['album'] . ') ' . '</a></div><hr /></li>';
        };
    };
    echo '</ul>';
}
else {

    // ================================ Default =========================================

    $query = "select * from music";
    $result = mysqli_query($link, $query) or die(mysqli_error());
    echo '<ul id="sortable1" class="connected">';
    while ($info = mysqli_fetch_array($result)) {
        echo '<li><div class="ui360"><button type="button" class="addtoplaylist" >Add</button><a href="' . $info['path'] . '"> ' . $info['artist'] . ' - ' . $info['title'] . ' (' . $info['album'] . ') ' . '</a></div><hr /></li>';
    };
    echo '</ul>';
};
?>

HTML：

<div id="browsemusic">
    <?php include ( 'php/searchbrowselist.php'); ?>
</div>

所以，php文件从ajax获取一些值并将结果加载到我的div中。我还需要它来执行我在头部加载的脚本：

<head>
    <script src="js/berniecode-animator.js"></script>
    <script src="js/soundmanager2.js"></script>
    <script src="js/360player.js"></script>
</head>

在ajax响应中从dom执行javascript

0 个答案: