Php MySQLi-返回二维数组

Php MySQLi-返回二维数组,php,mysqli,Php,Mysqli,有没有办法让MySQLi返回多维数组 假设我从数据库中选择博客帖子,每个帖子都有多个标签。据我所知,我会这样做: $r = $mysqli->query("SELECT * FROM posts"); while ($post = $r->fetch_assoc()) { //echo blog posts $t = $mysqli->query("SELECT * FROM blog_tags INNER JOIN tags ON tags.id = blog_

有没有办法让MySQLi返回多维数组

假设我从数据库中选择博客帖子,每个帖子都有多个标签。据我所知,我会这样做:

$r = $mysqli->query("SELECT * FROM posts");
while ($post = $r->fetch_assoc()) {
   //echo blog posts
    $t = $mysqli->query("SELECT * FROM blog_tags INNER JOIN tags ON tags.id = blog_tags.tag_id WHERE post_tags.post_id = ".$post['id']);
    while($tag = $t->fetch_assoc()){
        //echo tags
    }
}
但我更喜欢这样做:

$r = $mysqli->query("SELECT * FROM posts INNER JOIN post_tags ON posts.id = post_tags.post_id INNER JOIN tags ON tags.id = post_tags.tag_id");
while($post = $r->fetch_assoc()){
    //echo post
    foreach($post['tags'] as $tag){
        //echo tags
    }
}
因此,回报将是:

$posts = array(
    "id" => 1,
    "title" => "Blog post 1",
    "content" => "Lorem ipsum...",
    "tags" => 
        array(
            "tag1", 
            "tag2", 
            "tag3"
        ),
    "posted" => "1-1-2010 11:11:11"
);

您需要使用
GROUP BY
GROUP\u CONCAT
。我猜标记名字段名为
name

SELECT posts.*, GROUP_CONCAT(tags.name) as tagnames FROM posts
INNER JOIN post_tags ON posts.id = post_tags.post_id
INNER JOIN tags ON tags.id = post_tags.tag_id
GROUP BY posts.id

然后,当您得到结果时,您可以
分解(“,”,$t->标记名)
以获得一个标记数组。

我知道这不是很有帮助,但是如果您想要像这样查询和更新数据,您可能需要看看没有像MongoDB这样的sql解决方案。你在对象上插入和查询,结果和你发布的一模一样。看起来很有趣,一定会尝试的!