忽略用于将数据导出到php mysql中的csv文件的空列

Question

我正在使用一个脚本，该脚本可以从三个表（ playlist, songs, rate ）导出数据，并将其放入一个csv文件中，可以正常工作。

如果一个或多个列为空或具有空值，则它还会在csv文件中导出并显示为空列，如下所示：

和这个

和这个

因此，我希望如果列为空，则这些列不会在csv文件中导出。 我不知道该怎么办。

注意：我只希望空列不导出，不行

这是我在一个csv文件中导出三个表的代码。

$pre = $wpdb->prefix;

    $link = mysqli_connect($mysql_host,$mysql_user,$mysql_pass,$mysql_db) or die('Could not connect: '.mysqli_error());
    mysqli_select_db($link,$mysql_db) or die('Could not select database: '.$mysql_db);

    $query = "SELECT plist.*, psong.*, prate.* 
              FROM " . $pre . "hmp_songs As psong
              LEFT JOIN " . $pre . "hmp_playlists As plist
              On plist.playlist_name = psong.song_playlist 
              LEFT JOIN " . $pre . "hmp_rating As prate
              On psong.song_id = prate.rsong_id";

    $result = mysqli_query($link,$query) or die("Error executing query: ".mysqli_error());
    $row = mysqli_fetch_assoc($result);

    $line = "";
    $comma = "";

    foreach($row as $name => $value){
        $line .= $comma . '"' . str_replace('"', '""', $name) . '"';
        $comma = ",";
    }

    $line .= "\n";
    $out = $line;

    mysqli_data_seek($result, 0);
    while($row = mysqli_fetch_assoc($result)){
        $line = "";
        $comma = "";
        foreach($row as $value)
        {
            $line .= $comma . '"' . str_replace('"', '""', $value) . '"';
            $comma = ",";
        }
        $line .= "\n";
        $out .= $line;
    }

    $csv_file_name = 'HMP_'.date('Ymd_His').'.csv'; # CSV FILE NAME WILL BE table_name_yyyymmdd_hhmmss.csv
    header("Content-type: text/csv");
    header("Content-Disposition: attachment; filename=".$csv_file_name);
    header("Content-Description:File Transfer");
    header('Content-Transfer-Encoding: binary');
    header('Cache-Control: must-revalidate, post-check=0, pre-check=0');
    header('Pragma: public');
    header('Content-Type: application/octet-stream');
    echo __($out,"hmp");
    exit;

还有一件事：

完成之后，不导出空列，文件导入成功了吗？

Answer 1

好的，我感谢@vel弄清楚了。

我只是更改查询，它运行完美，并且也成功导入。

这是我的查询：

$query = "SELECT plist.playlist_id, plist.playlist_name, plist.playlist_shortcode, psong.song_id, psong.list_order,
              psong.song_playlist, psong.mp3, psong.ogg, psong.title, psong.buy, psong.buyy, psong.buyyy, psong.price, psong.cover, 
              psong.artist
              FROM " . $pre . "hmp_songs As psong
              LEFT JOIN " . $pre . "hmp_playlists As plist
              On plist.playlist_name = psong.song_playlist
              Where plist.playlist_id IS NOT NULL
              And plist.playlist_name IS NOT NULL
              And plist.playlist_shortcode IS NOT NULL
              And psong.song_id IS NOT NULL
              And psong.list_order IS NOT NULL
              And psong.song_playlist IS NOT NULL
              And psong.mp3 IS NOT NULL
              And psong.ogg IS NOT NULL
              And psong.title IS NOT NULL
              And psong.buy IS NOT NULL
              And psong.buyy IS NOT NULL
              And psong.buyyy IS NOT NULL
              And psong.price IS NOT NULL
              And psong.cover IS NOT NULL
              And psong.artist IS NOT NULL";

Answer 2

我认为array_filter可能在这里有用。

$row = array_filter( mysqli_fetch_assoc( $result ) );

并在循环中

while( $row = array_filter( mysqli_fetch_assoc( $result ) ) ){
    /* do stuff */
}