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如何使用php / mysql从对象/数组中获取正确的信息?

[英]How do I get the correct information out of my objects/ array with php / mysql?

I'm not a expert in programming but I want to set my old website over to new design and herefor I want to re-use my old code (slightly adjusted). 我不是编程专家,但是我想将旧网站设置为新设计,因此我想重用旧代码(略作调整)。

With the following function I get all artist information and guitar songs written. 使用以下功能,我可以编写所有艺术家信息和吉他歌曲。 To display in a table. 显示在表格中。

but the artist information (aArtist) is no problem but the song information is not working correctly. 但是艺术家信息(aArtist)没问题,但是歌曲信息无法正常工作。 I think I'm not far away but what should/could i try to make it work. 我想我并不遥远,但是我应该/应该设法使它起作用。

When printing (var_dump) the $aArtist I do get all information, its now hot to manage this. 当打印(var_dump)$ aArtist时,我确实获得了所有信息,现在管理该信息非常热门。

Thanks for all suggestions. 感谢您的所有建议。 Jan 一月

    function getArtistBySeoName($sSeoName,$sSortColumn,$sSortDirection,$sWhere)
{
    $sWhereAdditions = "";

    if($sWhereAdditions == "" && $sWhere != "")
    {
        $sWhereAdditions = " WHERE ".$sWhere;
    }
    $sSeoName = str_replace("'", "", $sSeoName);
    $sQuery = "SELECT * FROM artist a
    LEFT JOIN artist_info ai ON a.artist_id = ai.artist_info_artist_id
    WHERE artist_seoname = '".$sSeoName."'
    ".$sWhereAdditions." ";

    $oQueryResult = query($sQuery);

    $oArtist = $oQueryResult->fetch_assoc();

    $aSongs = array();
    $sQuery2 = "SELECT s.song_id,s.song_name,a.*,aa.*,st.*,ss.*,d.* FROM song s
    LEFT JOIN song_type st ON s.song_song_type_id = st.song_type_id
    LEFT JOIN song_style ss ON s.song_style_id = ss.song_style_id
    LEFT JOIN song_status sst ON sst.song_id   = s.song_id
    LEFT JOIN artist_album aa ON s.song_album_id = aa.artist_album_id
    LEFT JOIN download d ON s.song_id = d.download_song_id  inner join artist a on a.artist_id=s.song_artist_id
    WHERE s.song_artist_id = ".$oArtist[artist_id]."  and  (sst.status=1 or sst.status=2 or sst.status=3)  ORDER BY ".$sSortColumn." ".$sSortDirection;
    $oQueryResult2 = query($sQuery2);
    $i=0;

    while($oSong = $oQueryResult2->fetch_assoc())
    {
        $i++;
        $aSongs[]=$oSong;
    }

    $oArtist[songs] = $aSongs;

    $aAlbums = array();
    $sQuery3 = "SELECT * FROM artist_album aa
    INNER JOIN song s ON s.song_album_id = aa.artist_album_id
    WHERE s.song_artist_id = ".$oArtist[artist_id]." GROUP BY aa.artist_album_id";
    $oQueryResult3 = query($sQuery3);
    while($oAlbum = $oQueryResult3->fetch_assoc())
    {
        $aAlbums[]=$oAlbum;
    }
    $oArtist[albums] = $aAlbums;

    $aImages = array();
    $sQuery3 = "SELECT * FROM artist_images ai
    WHERE ai.artist_image_artist_id = ".$oArtist[artist_id]." ";
    $oQueryResult3 = query($sQuery3);
    while($oImage = $oQueryResult3->fetch_assoc())
    {
        $aImages[]=$oImage;
    }
    $oArtist[images] = $aImages;

    $oArtist[songscount] = $i;

    return $oArtist;
}
?>


<? 
$sSeoName = $route[2];
$sSortColumn ="song_name";
$sSortDirection="ASC"; 
$sWhere = ""; 

$aArtist = getArtistBySeoName($sSeoName,$sSortColumn,$sSortDirection,$sWhere); 

var_dump($aArtist);

Below here the code where I am trying to print the values. 在下面的代码中,我尝试打印值。 This is where it goes wrong probably. 这可能是哪里出错了。

                <?

            foreach($aArtist as $oSong)
            {

                //print $aArtist[song_name];

                switch($oSong[song_type_name])
                {
                    case "CRD":
                        $sPagename = CHORDSPAGE;
                    break;
                    case "TAB":
                        $sPagename = TABPAGE;
                    break;
                    case "BTAB":
                        $sPagename = BTABPAGE;
                    break;
                    case "LYRIC":
                        $sPagename = LYRICSPAGE;
                    break;
                    case "PTB":
                        $sPagename = PTBPAGE;
                    break;
                    case "GPR":
                        $sPagename = GPRPAGE;
                    break;
                } 

?> ?>

            <tr>
                    <td><span itemprop="name"><a href="/<?=$sPagename?>/<?=$oSong[artist_seoname]?>/<?=$oSong[song_seoname]?>"><? echo ucwords($oSong->song_name);?></a><span><meta itemprop="url" content ="http://www.gitaartabs.nl/<?=$sPagename?>/<?=$oSong[artist_seoname]?>/<?=$oSong[song_seoname]?>"><meta itemprop="duration" content="PT2M30S"></td>
                    <td class=""><?=ucfirst($oSong[song_type_name])?></td>
                    <td class="hidden-sm"><?=ucfirst($oSong[artist_album_name])?></td>
                    <td class="hidden-sm"><?=ucfirst($oSong[artist_album_year])?></td>
                    <td class="hidden-sm"><?=ucfirst($oSong[song_style_name])?></td>
                    <td><em><?=ucfirst($oSong[download_amount])?></em></td>
                    <td><i class="fa fa-star"></i><i class="fa fa-star"></i><i class="fa fa-star"></i><i class="fa fa-star"></i><i class="fa fa-star-half-o"></i></td>
                </tr>
            </tbody>
        </table>
        <!--End Basic Table-->  

I fixed this by using a foreach(song as songval) { } code 我通过使用foreach(歌曲作为Songval){}代码来解决此问题

thanks all for your help 感谢你的帮助

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