id - 电子邮件 - 用户名

1      | email |   bill

表2 - 图像

图片ID | id | cover_image

1        |     1    |   image1.jpg
2        |     1    |   image2.jpg
3        |     1    |   image3.jpg
4        |     1    |   image4.jpg
5        |     1    |   image5.jpg

我有一切显示，我不确定是否能够实现多个图像列表。

如果用户ID在上传图片时为1，则还会插入1

的ID

所以，如果用户有5张图片，我希望它显示为

<img src="images/gallery/$cover_img"></a>
<img src="images/gallery/$cover_img"></a>
<img src="images/gallery/$cover_img"></a>
<img src="images/gallery/$cover_img"></a>
<img src="images/gallery/$cover_img"></a>

但是现在它只会显示1张图片

$db = dbconnect();
$stmt = $db->prepare('SELECT * FROM `users` LEFT JOIN `images` ON images.id = users.id GROUP BY users.id');
$stmt->execute();
$result = $stmt->get_result();

Answer 1

无需使用join和group by query，您也可以使用foreach循环来获取值。

<?php  
 try {
        $sql1='select * from  users' ;
        foreach ($dbh->query($sql1) as $row)
             {
                $uid=$row['id']; 
                    $sql2='select * from images where id='.$uid.'' ;
                        foreach ($dbh->query($sql2) as $row2)
                         {
                             $images=$row2['cover_image'];
                         }
             }
               $dbh = null;
            } catch (Exception $ex) {
                echo "Some Exception Occured " . $ex;
            }
            ?>

Answer 2

您应该像下面一样更新您的查询，以获取您想要的正确数据。

以下是获取数据的MySQL query

示例1：获取所有用户图片

SELECT
 user.email,
 user.bill,
 images.id,
 images.cover_image
FROM user
INNER JOIN images ON user.id=images.id
ORDER BY images.id ASC

示例2：获取单个用户图片

SELECT
 user.email,
 user.bill,
 images.id,
 images.cover_image
FROM user
INNER JOIN images ON user.id=images.id
WHERE user.id= 1 #Here 1 is the user ID

以下是使用mysqli
的示例

users 表

+----------+-----------------+----------+
|    id    |      email      |  bill    |
+----------+-----------------+----------+
|    1     | email@email.com |  5000    |
+----------+-----------------+----------+

images 表

+----------+-----------------+----------------+
|    id    |      user_id    |  cover_image   |
+----------+-----------------+----------------+
|    1     |         1       |  image1.jpg    |
+----------+-----------------+----------------+
|    2     |         1       |  image2.jpg    |
+----------+-----------------+----------------+
|    ...   |        ...      |      ...       |
+----------+-----------------+----------------+

PHP and MySQL 使用mysqli

的代码

#$db = dbconnect();
$db = new mysqli("HOSTNAME", "USER", "PASSWORD", "DATABASE_NAME");

if($db->connect_error){
  die("$db->connect_errno: $db->connect_error");
}

$sql = '
    SELECT
        users.email,
        users.bill,
        images.user_id,
        images.cover_image
    FROM users
    INNER JOIN images ON users.id=images.user_id
    ORDER BY images.id ASC';

 $stmt   = $db->prepare($sql);
 $stmt->execute();
 $result = $stmt->get_result();

Display Images 来自$result

<?php while ($row = $result->fetch_array(MYSQLI_ASSOC)){ #Get every row as an Associative array?>
    <img src="images/gallery/<?php echo $row['cover_image'] ?>" alt="<?php echo $row['cover_image'] ?>"></a>
<?php } ?>

Answer 3

要实现这一点，你需要用你的mysql查询替换LEFT JOIN和RIGHT JOIN。

对于每个循环或数组

id - 电子邮件 - 用户名

图片ID | id | cover_image

3 个答案: