mysqli獲取記錄

這是關於一個職位Show three images from each user，其中有一個很好的解決方案，我遇到了同樣的問題。mysqli獲取記錄

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>

我想知道如何也可以在此查詢中獲取pictureID？例如，如果我使用圖片進行投票，那麼我還需要獲取pictureID。 Supericy幫助後

答：（如果有人找相同）

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")) {

然後將它綁定

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

來源

2012-12-25 Awara Leo

假設圖片的ID列被命名爲「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

，然後綁定時的結果：

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

來源

2012-12-25 00:34:43 Supericy

聽起來不錯，但我接下來要它的變量綁定以及在bind_result？ –

哦，是的。 '$ stmt-> bind_result（$ user，$ pic1，$ pic1_id，$ pic2，$ pic2_id，$ pic3，$ pic3_id）;'（編輯我的答案） – Supericy

我試過了，但它給出了「Warning：mysqli_stmt :: bind_result（） [mysqli-stmt.bind-result]：綁定變量的數量與預編譯語句中的字段數量不匹配「 –

試試這個，

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

來源

2012-12-25 00:30:53 Mari

mysqli獲取記錄

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