结合的MySQL查询

Question

我在DB两个不同的表，这里的相关字段：

表图片：

CREATE TABLE `images` (
    `image_id` int(4) NOT NULL AUTO_INCREMENT, 
    `project_id` int(4) NOT NULL, 
    `user_id` int(4) NOT NULL, 
    `image_name` varchar(255) COLLATE utf8_unicode_ci NOT NULL, 
    `image_description` varchar(255) COLLATE utf8_unicode_ci NOT NULL, 
    `date_created` date NOT NULL, 
    `link_to_file` varchar(255) COLLATE utf8_unicode_ci NOT NULL, 
    `link_to_thumbnail` varchar(255) COLLATE utf8_unicode_ci NOT NULL, 
    `given_name` varchar(255) COLLATE utf8_unicode_ci NOT NULL, 
    `note` varchar(255) COLLATE utf8_unicode_ci NOT NULL, 
    PRIMARY KEY (`image_id`) 
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci AUTO_INCREMENT=51 ; 

并表项目：

CREATE TABLE `projects` (
    `project_id` int(4) NOT NULL AUTO_INCREMENT, 
    `user_id` int(4) NOT NULL, 
    `project_name` varchar(255) COLLATE utf8_unicode_ci NOT NULL, 
    `project_description` varchar(255) COLLATE utf8_unicode_ci NOT NULL, 
    `date_created` date NOT NULL, 
    `date_last_edited` date NOT NULL, 
    `shared` int(1) NOT NULL, 
    `password` varchar(255) COLLATE utf8_unicode_ci NOT NULL, 
    PRIMARY KEY (`project_id`) 
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci AUTO_INCREMENT=25 ; 

我想展示在变量$ content中，每个项目中最老的图像作为该项目页面的链接，我不知道应该如何构建mysql查询。你能帮我解决这个问题吗？我已经尝试了几个if和while语句，但结果已经完全失败，而我处于我的（非常有限）知识的末尾。我都快要跳出窗口...

所以我想用

<a href="index.php?page=projects&id='.$projectid.'"> 
    <img src="oldest_photo_of_project_x" /> 
</a> 
<a href="index.php?page=projects&id='.$projectid.'"> 
    <img src="oldest_photo_of_project_y" /> 
</a> 
<a href="index.php?page=projects&id='.$projectid.'"> 
    <img src="oldest_photo_of_project_z" /> 
</a> 

UPDATE1落得：

为了澄清，我想结合：

"SELECT * FROM projects WHERE user_id='$UserID' ORDER BY project_id DESC" 

也许是这样的：

$query = "SELECT images.project_id, projects.project_name ". 
"FROM images, projects ". 
"WHERE images.project_id = projects.project_id"; 

Answer 1

没有测试的错误，但我会做这样的事情：

$result = mysql_query("SELECT DISTINCT 
    `projects`.`project_id` AS `project`, 
    `images`.`link_to_file` AS `filepath` 
FROM 
    `projects`, 
    `images` 
WHERE 
    `projects`.`project_id` = `images`.`project_id` 
ORDER BY 
    `images`.`date_created` DESC"); 

while ($resultLoop = mysql_fetch_array($result)) { 
    $str .= '<a href="index.php?page=projects&id=' . $resultLoop["project"] . '"> 
     <img src="' . $resultLoop["filepath"] . '" /> 
    </a>'; 
} 


echo $str; 

Answer 2

像这样的东西会从数据库中提取图像上的数据。

SELECT * FROM `images` WHERE `project_id`="x" ORDER BY `date_created` ASC LIMIT 1,1; 

这将为ID为'x'的项目提取最早的图像。你可以把它们连起来，如下所示：

SELECT * FROM `images` WHERE `project_id`="x" ORDER BY `date_created` ASC LIMIT 1,1; 
SELECT * FROM `images` WHERE `project_id`="y" ORDER BY `date_created` ASC LIMIT 1,1; 
SELECT * FROM `images` WHERE `project_id`="z" ORDER BY `date_created` ASC LIMIT 1,1; 

如果您使用的是PHP，你可以使用mysqli_result :: fetch_array得到它应该包含所有3次结果的查询结果。

Answer 3

$dbh = new PDO($DSN, $USERNAME, $PASSWORD); 
$qry = $dbh->prepare(' 
    SELECT project_id, link_to_thumbnail 
    FROM  images NATURAL JOIN (
    SELECT project_id, MIN(date_created) AS date_created 
    FROM  images 
    GROUP BY project_id 
    WHERE user_id = ? 
) AS t 
    ORDER BY project_id DESC 
'); 
$qry->bindValue(1, $UserID); 
$qry->execute(); 

while ($row = $qry->fetch()) echo " 
    <a href=\"index.php?page=projects&id=$row[project_id]\"> 
    <img src=\"$row[link_to_thumbnail]\"/> 
    </a> 
";