PHP 无法在满足条件后回显文本

Question

我没有从下面的代码中得到所需的结果。一切正常，但有一个主要问题：当数据库中没有歌曲时，它仅通过后退按钮显示空结果。它无法显示回声“抱歉！” 我们的数据库中没有这首歌。我不知道错误在哪里。所以请帮忙。提前致谢！！！<?php// Php code that fetches audio from the database in server and shows the audio files with singers available for the submitted data// In submission page DROPDOWN consists of Playlist name and TEXTBOX allows to type the song number for that playlist.// Standard format for the audio file stored in the databse is Songnumber-Playlistname-Singer's Shortname.mp3// MP3 files will be inside the AUDIO folder and this PHP code runs from the root folder where there is index.html file for data submission.// Valid song Numbers of Each Playlists that user choose$validsongnumbers = [    'Rock' => 3,     'Pop' => 5,     'Jazz' => 6];// Data captured from dropdown submitted by a user in homepage $PlaylistName = $_POST['Dropdown'];$songNumber = $_POST['songnumber'];// Check the playlist existsif (!array_key_exists($PlaylistName, $validsongnumbers)) {    echo 'Invalid playlist provided.';    exit;}// Check the song number is not greater than what is allowedif ((int)$songNumber > $validsongnumbers[$PlaylistName]) {    echo  'Invalid song number provided.';    exit;}

Answer 1

首先，你将“if if”加倍：

if if(count($files) < 0

其次，文件数量永远不能为负。您应该比较 number 是否等于 0（零），而不是小于 0。

更新：

还是错误的代码：

if ($count=0)

这不是比较，而是赋值。您给出的 $count 值为 0。为了进行比较，您必须使用：

if ($count == 0)