我试图通过使用数据库通过MYSQL和PHP创建XML文件来制作电视指南。到目前为止我已经取得了很好的成功,并且已经设法输出每天的各种频道。如何使用PHP和带条件嵌套的MYSQL创建XML
问题是当我尝试带入各种程序时。在数据库中,每个程序都被分配一天并显示频道。然而,正在生成的XML将每个节目放在每个频道的每一天。
这是我到目前为止有:
$xml = new DOMDocument("1.0", "UTF-8");
$xml->formatOutput = TRUE;
$day_query = "SELECT day_name FROM day;";
$dq_result = mysql_query($day_query) or die(mysql_error());
$tvguide = $xml->createElement("guide");
$tvguide = $xml->appendChild($tvguide);
if($dq_result)
while($dayrow = mysql_fetch_row($dq_result)){
$day = $xml->createElement("day");
$day->setAttribute("name", $dayrow[0]);
$day = $tvguide->appendChild($day);
$channel_query = "SELECT channel_ID, channel_Name FROM channel ORDER BY channel_Name ASC;";
$cq_result = mysql_query($channel_query) or die(mysql_error());
if($cq_result)
while($channelrow = mysql_fetch_row($cq_result)){
$channel = $xml->createElement("channel");
$channel->setAttribute("id", $channelrow[0]);
$channel = $day->appendChild($channel);
$channel_name = $xml->createElement("channel_name");
$channelNameText = $xml->createTextnode($channelrow[1]);
$channel_name->appendChild($channelNameText);
$channel_name = $channel->appendChild($channel_name);
$show_query = "SELECT programme_ID, programme_Name, start_Time, duration, description, day_Name, channel_Name FROM programme ORDER BY start_Time ASC;";
$sq_result = mysql_query($show_query) or die(mysql_error());
if($sq_result)
while($showrow = mysql_fetch_row($sq_result)){
$show = $xml->createElement("show");
$show = $channel->appendChild($show);
$show_name = $xml->createElement('show_name');
$show_name->setAttribute("id", $showrow[0]);
$showNameText = $xml->createTextnode($showrow[1]);
$show_name->appendChild($showNameText);
$show->appendChild($show_name);
}
}
}
基本上,我想知道我怎么可以嵌套所有其DAY_NAME等于星期一,例如,“星期一” XML标签中显示。谢谢。
非常好,非常感谢你的帮助=) – user2950001 2014-12-03 14:27:16