在我的 MySQL 数据库上有 3 个不同的表:专辑、单曲、即将到来的。我想通过 php 嵌入所有三个。只有我嵌入的第一个表适用于以下代码,另一个不适用:
<div id="album_releases" class="animated bounceIn">
<div id="headline">Latest Favorite Album Releases</div>
<div id="frame" class="frame">
<ul class="slidee">
<li><div id="datebox"><span style="padding-top:50px;">Oct<br />2013</span></div></li>
<?php
$sql = "SELECT artist, releasename, label, date, img, spotify, nas, itunes FROM album ORDER BY id DESC LIMIT 40";
$db_erg = $db_link->query( $sql );
function ausgabe($daten) {
echo "<li><div id='blurbox'><div class='blur img'><div class='img-wrap'><div class='img-info'>";
echo "". $daten->artist ."<br />";
echo "". $daten->releasename ."<br />";
echo "". $daten->label ."<br />";
echo "". $daten->date ."<br />";
echo "</div>";
echo "<img class='hover' width='168' height='168' src=". $daten->img .">";
echo "</div></div></div><div id='caption'>";
echo "<img src='images/spotifyicon.png' style='float:left;'><a href='". $daten->spotify ."'>Listen on Spotify</a><br />";
echo "<div id='element_sep'>-</div>";
echo "<img src='images/servericon.png' style='float:left;'><a href='". $daten->nas ."'>Listen on toegrNAS</a><br />";
echo "<div id='element_sep'>-</div>";
echo "<img src='images/itunesicon.png' style='float:left;'><a href='". $daten->itunes ."'>Listen/Buy on iTunes</a>";
echo "</div></li>";
}
while($daten = $db_erg->fetch_object()) {
ausgabe($daten);
}
?>
</ul>
</div>
<div id="overlay_left">-</div>
<div id="overlay_right">-</div>
</div>
<div id="single_releases" class="animated bounceIn">
<div id="headline">Latest Favorite Single Releases</div>
<div id="frame" class="frame">
<ul class="slidee">
<li><div id="datebox"><span style="padding-top:50px;">Oct<br />2013</span></div></li>
<?php
$sql = "SELECT artist, releasename, label, date, img, spotify, nas, itunes FROM single ORDER BY id DESC LIMIT 40";
$db_erg = $db_link->query( $sql );
function ausgabe($daten) {
echo "<li><div id='blurbox'><div class='blur img'><div class='img-wrap'><div class='img-info'>";
echo "". $daten->artist ."<br />";
echo "". $daten->releasename ."<br />";
echo "". $daten->label ."<br />";
echo "". $daten->date ."<br />";
echo "</div>";
echo "<img class='hover' width='168' height='168' src=". $daten->img .">";
echo "</div></div></div><div id='caption'>";
echo "<img src='images/spotifyicon.png' style='float:left;'><a href='". $daten->spotify ."'>Listen on Spotify</a><br />";
echo "<div id='element_sep'>-</div>";
echo "<img src='images/servericon.png' style='float:left;'><a href='". $daten->nas ."'>Listen on toegrNAS</a><br />";
echo "<div id='element_sep'>-</div>";
echo "<img src='images/itunesicon.png' style='float:left;'><a href='". $daten->itunes ."'>Listen/Buy on iTunes</a>";
echo "</div></li>";
}
while($daten = $db_erg->fetch_object()) {
ausgabe($daten);
}
?>
</ul>
</div>
<div id="overlay_left">-</div>
<div id="overlay_right">-</div>
</div>
<div id="upcoming_releases" class="animated bounceIn">
<div id="headline">Upcoming Releases</div>
<div class="content">
<div id="release_element">
<?php
require_once ('mysql-config.php');
$db_link = mysqli_connect (
MYSQL_HOST,
MYSQL_BENUTZER,
MYSQL_KENNWORT,
MYSQL_DATENBANK
);
$sql = "SELECT date, artist, releasename, itunes, ical, directdl FROM upcoming ORDER BY id DESC LIMIT 40";
$db_erg = $db_link->query( $sql );
function ausgabe($daten) {
echo "<img src='images/icon_cd.png'>";
echo "<span class='release_bold'>". $daten->date .": ". $daten->artist ." - ". $daten->releasename ."</span><br />";
echo "<span class='release_grey'><a href='". $daten->itunes ."'>iTunes (Pre-Order)</a> - <a href='". $daten->ical ."'>Create iCal-Event</a> - <a href='". $daten->directdl ."'>Download</a></span><br />";
}
while($daten = $db_erg->fetch_object()) {
ausgabe($daten);
}
?>
当我尝试嵌入其他两个表时,页面只显示一个白屏。问题是,如何通过 php 在不同位置嵌入三个不同的 mysql 表?
谢谢
项目链接: http: //toegraphics.de/music/