我有一个图片表,其中用户图片连同他们的 ID 和图片物理链接一起保存。
userID | picture
1 | picnameLink
1 | picnameLink
2 | picnameLink
1 | picnameLink
2 | picnameLink
3 | picnameLink
现在,我想在 jquery 图片库 block 中显示最多 3 张图片,其中一个 block 应显示来自同一用户的所有 3 张图片,如果用户的图片少于 3 张,则不应显示任何图片文本。
我尝试使用 group by mysql 查询,但没有得到想要的结果。我必须使用两个循环吗?
--为 fthiella 编辑--这是代码
$query = "SELECT * FROM pictures GROUP BY userID";
$result = mysql_query($query);
while($row = mysql_fetch_array($result)){
$image_array[] = $row['picLink'];
$id_array[] = $row['pic_id'];
}
$num_images_to_display = 3; /* MODIFY TO REFLECT NUMBER OF IMAGES TO SHOW PER SCREEN */
$num_images = count($image_array);
$image_path = "../images/"; /* MODIFY TO REFLECT THE PATH TO YOUR IMAGES */
$y = $num_images_to_display;
if(!isset($_GET['first'])){
$first = 0;
}else{
$first = (int) $_GET['first'];
}
$x = $num_images - 1;
$z = $x - $y;
if($first>$z) {
$first = $z;
}
$last = $first + $num_images_to_display;
这里是 HTML 区域:
<div style="position:absolute; top:50px; left:100px; width:800px; text-align: center;">
<?PHP
$i = $first;
while($i<$last) { $showme = $image_path . $image_array[$i]; ?>
<?php if($image_array[$i]!="") { ?><img src="<?PHP echo $showme; ?>" width="176px" height="197px"><?php } else { ?><img src="../image/no_image.jpg" width="176px" height="197px"><?PHP } ?>
$prev = $first-1;
$next = $first +1;
if($prev<0){ $prev = 0; }
?>
</div>
此查询的结果按组显示图片,但我希望每个用户最多三张图片,如果用户的图片少于三张,则不会显示任何图片。
最佳答案
我不知道是否有更好的解决方案,但我认为你可以使用这个:
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
这将为每个用户选择三张图片。如果用户拥有的图片少于三张,它会显示空值,如果有更多,它会在所有图片中选择三张。
另一种方法是使用变量在不同的行中显示三个图像:
SELECT userid, picture
FROM (
SELECT
userid,
picture,
case when @prec_id=userid then @row:=@row+1 else @row:=1 end as row,
@prec_id:=userid
FROM
`pictures`,
(SELECT @prec_id:=0, @row:=0) s
ORDER BY userid) s
WHERE row<=3
编辑:要一次为每个用户显示三张图片,我会使用我的第一个查询,我将从这样的代码开始:
<?php
$mysqli = new mysqli("localhost", "username", "password", "test");
$image_path = "../images/";
$no_image = "../image/no_image.jpg";
if(!isset($_GET['first'])){
$first = 0;
} else {
$first = (int) $_GET['first'];
}
if ($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>
(它已得到改进,但您可以从它开始。我使用的是 mysqli 而不是 mysql)
关于php - 显示每个用户的三张图片,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13980078/