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：

1. 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 exists

if (!array_key_exists($PlaylistName, $validsongnumbers)) {

    echo 'Invalid playlist provided.';

    exit;

}

// Check the song number is not greater than what is allowed

if ((int)$songNumber > $validsongnumbers[$PlaylistName]) {

    echo  'Invalid song number provided.';

    exit;

}

Answer 1

首先，你将“if if”加倍：if&nbsp;if(count($files)&nbsp;<&nbsp;0其次，文件数量永远不能为负。您应该比较 number 是否等于 0（零），而不是小于 0。更新：还是错误的代码：if&nbsp;($count=0)这不是比较，而是赋值。您给出的 $count 值为 0。为了进行比较，您必须使用：if&nbsp;($count&nbsp;==&nbsp;0)