我有两个表
- 照片：存储所有照片信息
- 库：存储库的标题和说明
我需要回显到一个新的页面画廊描述，然后所有的照片应该在那里。
但我不断重复标题和描述，因为它在同一个中
$row = mysql_fetch_array($result)...
那么我还需要一组照片吗？
有人帮忙吗？还是我太含糊了。。。。
$a="SELECT * from gallery where gallery_category=".$gallery_category;
$result = mysql_query($a,$c);
while($row = mysql_fetch_array($result)) {
echo $row['gallery_name'].'<br>'.$row['gallery_description'].'<br>';
$sql="SELECT * FROM photos WHERE gallery_id =".$gallery_id." ORDER BY photos_filename";
$result2 = mysql_query($sql,$c);
while($row = mysql_fetch_array($result2)) {
echo'<a rel="example_group" href="../galleries/images/'.$row['gallery_id'].'/'.$row['photos_filename'].'" width="130" height="100"><img src="../galleries/images/'.$row['gallery_id'].'/'.$row['photos_filename'].'" height="150px" alt=""/></a>';
}
}
实际上我会把它发布在这里，这样我就可以使用一些代码格式了。
选项1-循环内循环
$headerquery = $db->query("SELECT * FROM tbl1");
while ($headers= $db->fetchNextObject($headerquery )) {
echo "<table><tr><th>".$headers->GalleryName."</th></tr>"; // open table create header
$detailquery = $db->query("SELECT * FROM tbl2 WHERE GalleryID=".$headers->ID);
while ($details= $db->fetchNextObject($detailquery )) {
echo "<tr><td>".$details->Photo."</td></tr>"; //loop through 2nd table and spit out photos
}
echo "</table>"; //close table
}
选项2-使用选择器加入查询
$galleryheaderid = 0;
$query = $db->query("SELECT * FROM tbl1 INNER JOIN tbl2 on tbl1.ID=tbl2.GalleryID");
while ($details = $db->fetchNextObject($query )) {
echo "<table>";
if ($galleryheaderid!=$details->ID) { //spit out header row if id's don't match
echo "<tr><th>".$details ->GalleryName."</th></tr>"; //create header
$galleryheaderid = $details->ID; //set header id to gallery id so we don't show header a 2nd time
}
echo "<tr><td>".$details->Photo."</td></tr>"; //spit out gallery information even on first row since all rows include photo information
echo "</table>"; //close table
}
这两种方法中的任何一种都应该有效，显然它们需要正确完成