2

这是关于一个帖子显示来自每个用户的三张图片，其中有一个很好的解决方案来解决我遇到的相同问题。

if ($stmt = $mysqli->prepare("SELECT p1.userID, p1.picture as pic1, p2.picture as pic2, p3.picture as pic3
FROM
  pictures p1 left join pictures p2
  on p1.userID=p2.userID and p1.picture<>p2.picture
  left join pictures p3
  on p1.userID=p3.userID and p1.picture<>p3.picture and p2.picture<>p3.picture
GROUP BY p1.userID
LIMIT ?,1")) {

    $stmt->bind_param("i", $first); 
    $stmt->execute();
    $stmt->bind_result($user, $pic1, $pic2, $pic3);
    $stmt->fetch();
    $stmt->close();
}
$mysqli->close();
?>
<div style="position:absolute; top:50px; left:100px; width:800px; text-align: center;">
  <img src="<?PHP echo (isset($pic1) ? $image_path.$pic1 : $no_image); ?>" width="176px" height="197px">
  <img src="<?PHP echo (isset($pic2) ? $image_path.$pic2 : $no_image); ?>" width="176px" height="197px">
  <img src="<?PHP echo (isset($pic3) ? $image_path.$pic3 : $no_image); ?>" width="176px" height="197px">
</div>

我想知道如何在此查询中也获取图片ID？例如，如果我使用图片进行投票，那么我也必须获得图片ID。

在 Supericy 帮助后回答：（如果有人在寻找相同的东西）

if ($stmt = $mysqli->prepare("SELECT p1.userID, p1.picture as pic1, p1.pictureID as pic1id, p2.picture as pic2, p2.pictureID as pic2id, p3.picture as pic3, p3.pictureID as pic3id
FROM
  pictures p1 left join pictures p2
  on p1.userID=p2.userID and p1.picture<>p2.picture
  left join pictures p3
  on p1.userID=p3.userID and p1.picture<>p3.picture and p2.picture<>p3.picture
GROUP BY p1.userID
LIMIT ?,1")) {

然后绑定

$stmt->bind_result($user, $pic1, $pic1id, $pic2, $pic2id, $pic3, $pic3id);
4

2 回答 2

1

试试这个，

SELECT p1.userID,coalesce(p1.pictureID,p2.pictureID,p3.pictureID) pictureID, 
       coalesce(p1.picture,p2.picture,p3.picture) as pic
FROM
  pictures p1 left join pictures p2
  on p1.userID=p2.userID and p1.picture<>p2.picture
  left join pictures p3
  on p1.userID=p3.userID and p1.picture<>p3.picture and p2.picture<>p3.picture
GROUP BY p1.userID,pictureID,pic
LIMIT ?,1
于 2012-12-25T00:30:53.933 回答
0

假设图片的 ID 列名为“pictureID”：

SELECT p1.userID, p1.picture as pic1, p1.pictureID as pic1id, p2.picture as pic2, p2.pictureID as pic2id, p3.picture as pic3, p3.pictureID as pic3id
FROM
  pictures p1 left join pictures p2
  on p1.userID=p2.userID and p1.picture<>p2.picture
  left join pictures p3
  on p1.userID=p3.userID and p1.picture<>p3.picture and p2.picture<>p3.picture
GROUP BY p1.userID
LIMIT ?,1

然后在绑定结果时：

$stmt->bind_result($user, $pic1, $pic1id, $pic2, $pic2id, $pic3, $pic3id);
于 2012-12-25T00:34:43.077 回答