SQL查询：2个表，但希望从一个只有一行从其他

Question

所有的信息，我有2个表

• 照片：存储所有照片信息
• 库：存储标题和描述关于画廊

  ---

我需要回显一个新页面的图库描述，然后所有的照片应该在那里。

，但我不断收到标题和描述重复的，因为它在同

$row = mysql_fetch_array($result)... 

然后有更多的一组该库中我也需要照片？

任何帮助还是我是含糊....

$a="SELECT * from gallery where gallery_category=".$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>'; 
    } 
} 

Answer 1

其实我会张贴在这里，所以我可以使用一些代码格式化。

选项1 - 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 
} 

方案2中环 - 连接的查询与选择

$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 
} 

像任一的东西应该工作显然他们需要正确填写