Question

我一直都没有，也没有什么对我有用所以我不得不问社区他们是否有任何东西给我。

所以，我有一个下拉菜单和一个开关（复选框），如果我按下开关它应该加载下拉内容，让我们说＆＃34; Radio＆＃34;如果我将其切换回来，它应该将下拉内容切换回让我们说＆＃34;播放列表＆＃34;。

以下是代码：

<?PHP
// Read playlist & Radio
    $playlists =    explode("\n", shell_exec("/usr/bin/mpc lsplaylists"));
    array_pop($playlists); //remove empty last line

    $radioStations  = array();
    $radioStations =    explode("\n", shell_exec("/usr/bin/mpc ls Webradio"));
    $radioStations = str_replace("Webradio/","",$radioStations);
    $radioStations = str_replace(".pls","",$radioStations);
    array_pop($radioStations); //remove empty last line
?>

脚本：

<script>

        function checkSelectedScheduleType()
        {
            if (document.getElementById('radioSelected').checked) 
            {
                alert("radio selected fill dropdown with radio stations...");

                <?php 
                $playlist = $radioStations;
                ?>
            } 
            else 
            {
                <?php 
                $playlist = $playlists;
                ?>
                alert("Playlist selected fill dropdown with playlists...");
            }
        }


</script>

开关：

<td class="alarmvalue" style="padding:2px 15px;">
        <select name="add[alarmplaylist]" id="playlist_wrongstyle" class="form-control"  style="visibility:visible; background: transparent; width:250px;">
            <?php foreach ($playlist as $playlistss){echo "<option value=\"$playlistss\">$playlistss</option>\n";} ?>
        </select>
</td>

Answer 1

我发现的问题：

即使PHP是在JS＆＃34; IF＆＃34;声明，php仍然在没有调用JS函数的情况下运行，这很奇怪，所以select选择了最后一个php变量。

我的修复：

可能不是最好但是有效！我创建了另一个下拉列表，填充了我想要的选项，然后使Display =＆＃34; none＆＃34;然后当我点击它将触发我的功能＆＃34; onchange＆＃34;并使其显示=＆＃34;阻止＆＃34;互相隐藏，工作正常！

以下代码：

JS：

<script>
        function checkSelectedScheduleType()
        {
            if (document.getElementById('radioSelected').checked) 
            {
                document.getElementById("slplaylist").style.display="none";
                document.getElementById("slradio").style.display="block";
            } 
            else 
            {
                document.getElementById("slplaylist").style.display="block";
                document.getElementById("slradio").style.display="none";
            }
        }
</script>

HTML：

<td class="alarmvalue" style="padding:2px 15px;">
        <label class="switch-lightp2 well_" onchange="checkSelectedScheduleType()">
            <input id="radioSelected" name="" type="checkbox" value="">
            <span><span>Playlist</span><span>Radio</span></span><a class="btn btn-primary"></a>
        </label>
    </td>

    <td class="alarmcat">

    </tr>

    <td class="alarmvalue" id="slplaylist" style="padding:2px 15px; display:block;">
        <select name="add[alarmplaylist]" id="playlist_wrongstyle" class="form-control"  style="visibility:visible; background: transparent; width:250px;">
            <?php foreach ($playlists as $playlist){echo "<option value=\"$playlist\">$playlist</option>\n";} ?>
        </select>
    </td>

    <td class="alarmvalue" id="slradio" style="padding:2px 15px; display:none;">
        <select name="add[alarmplaylist]" id="playlist_wrongstyle" class="form-control"  style="visibility:visible; background: transparent; width:250px;">
            <?php foreach ($radioStations as $playlist){echo "<option value=\"$playlist\">$playlist</option>\n";} ?>
        </select>
    </td>

重新载入下拉框内容？

1 个答案: