2

これは、私が抱えていた同じ問題に対する優れた解決策がある各ユーザーからの 3 つの画像を表示する投稿に関するものです。

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>

このクエリで、pictureID も取得するにはどうすればよいのでしょうか。たとえば、投票に写真を使用する場合、pictureID も取得する必要があります。

Supericy ヘルプの後の回答: (誰かが同じものを探している場合)

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")) {

そして、それをバインドします

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

2 に答える 2

1

これを試して、

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 に答える
0

画像のID列の名前が「pictureID」であると仮定します。

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

そして、結果をバインドするとき：

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