[英]MySQL query to retrieve related data from another table
I'm not too experienced with SQL queries (most times I used query builder from frameworks, like CodeIgniter and Laravel).我对 SQL 查询不太熟悉(大多数时候我使用框架中的查询构建器,如 CodeIgniter 和 Laravel)。 Now, I need to retrieve data from a relational DB which have two tables: one for entity entries and other for complemental data of entities.现在,我需要从具有两个表的关系数据库中检索数据:一个用于实体条目,另一个用于实体的补充数据。 For example, see below:例如,见下文:
tbl_posts tbl_posts
id ID | name姓名 | slug蛞蝓 |
---|---|---|
1 1 | Lorem ipsum Lorem ipsum | lorem-ipsum lorem-ipsum |
2 2 | Testing post测试帖 | testing-post测试站 |
3 3 | Hello world你好世界 | hello-world你好世界 |
tbl_posts_meta tbl_posts_meta
id ID | post_id post_id | key钥匙 | value价值 |
---|---|---|---|
1 1 | 1 1 | first_name名 | John约翰 |
2 2 | 1 1 | last_name姓 | Doe能源部 |
3 3 | 1 1 | rating评分 | 5 5 |
4 4 | 2 2 | parent父母 | 1 1 |
5 5 | 2 2 | rating评分 | 3 3 |
6 6 | 3 3 | rating评分 | 4 4 |
In this example, I need to retrieve an array of objects in this format:在这个例子中,我需要以这种格式检索一个对象数组:
[
{
id: 1,
name: "Lorem ipsum",
slug: "lorem-ipsum",
data: {
first_name: "John",
last_name: "Doe",
rating: 5,
}
},
{
id: 2,
name: "Testing post",
slug: "testing-post",
data: {
parent: 1,
rating: 3,
}
},
{
id: 3,
name: "Hello World",
slug: "hello-world",
data: {
rating: 4,
}
},
]
I've tried using subquery to handle this, but I'm reveiving Cardinality violation: 1241 Operand should contain 1 column(s)
error.我已经尝试使用子查询来处理这个问题,但我正在恢复Cardinality violation: 1241 Operand should contain 1 column(s)
错误。 My query looks like this:我的查询如下所示:
SELECT *,(SELECT * FROM tbl_posts_meta WHERE "post_id" = "tbl_posts"."id") as data FROM tbl_posts;
I've already tried using JOIN, but the result looks more away from expected (I have one "key" property and one "value" property in result, containing the last found entry in tbl_posts_meta).我已经尝试过使用 JOIN,但结果看起来与预期相差甚远(结果中有一个“键”属性和一个“值”属性,包含 tbl_posts_meta 中最后找到的条目)。
SELECT * FROM tbl_posts INNER JOIN tbl_posts_meta ON tbl_posts_meta.post_id = tbl_posts.id;
Is there someway I can retrieve desired results with one query?有没有办法通过一个查询来检索所需的结果? I don't want to do this by applogic (like first retrieving data from tbl_posts and appending another query on "data" property, that returns all data from tbl_posts_meta), since this way may cause database overload.我不想通过 applogic 执行此操作(例如首先从 tbl_posts 检索数据并在“data”属性上附加另一个查询,该查询返回来自 tbl_posts_meta 的所有数据),因为这种方式可能会导致数据库过载。
Thanks!谢谢!
The code is not shown right in the comment so here it is for you to check, sorry I can't contribute more to the theoretical debate:该代码未在评论中正确显示,因此您可以在这里检查,抱歉,我无法为理论辩论做出更多贡献:
select
tp.id,
tp.name,
tp.slug,
(select group_concat(`key`, "=",value separator ';' ) from tbl_posts_meta where post_id= tp.id )as datas
from tbl_posts tp
You can follow these simple steps to get to your needed json:您可以按照以下简单步骤获取所需的 json:
tbl_posts_meta
JSON-like string by aggregating on your columns of interest ( tbl_posts.id
, tbl_posts.name
, tbl_posts.slug
) using a GROUP_CONCAT
aggregation function.通过使用GROUP_CONCAT
聚合函数聚合您感兴趣的列( tbl_posts.id
、 tbl_posts.name
、 tbl_posts.slug
)来生成类似tbl_posts_meta
的 JSON 字符串。 Since your keys can either be integers or strings, you need to do a quick check using the CAST(<col> AS UNSIGNED)
, which will return 0 if the string is not a number.由于您的键可以是整数或字符串,因此您需要使用CAST(<col> AS UNSIGNED)
进行快速检查,如果字符串不是数字,它将返回 0。JSON_OBJECT
function to transform each row into a JSON使用JSON_OBJECT
函数将每一行转换为 JSONJSON_ARRAYAGG
function to merge all jsons into a single json.使用JSON_ARRAYAGG
函数将所有 json 合并为一个 json。WITH cte AS(
SELECT p.id,
p.name,
p.slug,
CONCAT('{',
GROUP_CONCAT(
CONCAT('"',
m.key_,
'": ',
IF(CAST(m.value_ AS UNSIGNED)=0,
CONCAT('"', m.value_, '"'),
CAST(m.value_ AS UNSIGNED)) )
ORDER BY m.id ),
'}') AS data_
FROM tbl_posts p
INNER JOIN tbl_posts_meta m
ON p.id = m.post_id
GROUP BY p.id,
p.name,
p.slug
)
SELECT JSON_ARRAYAGG(
JSON_OBJECT("id", id,
"name", name,
"slug", slug,
"data", CAST(data_ AS JSON)))
FROM cte
After some research, testing and reading the queries you posted here, I've finally reached the result I needed.经过一些研究、测试和阅读您在此处发布的查询,我终于达到了我需要的结果。 The final query is structured this way:最终查询的结构如下:
SELECT *, (
SELECT JSON_ARRAYAGG(
JSON_OBJECT(
key, value
)
) FROM tbl_posts_meta WHERE post_id = post.id
) AS data
FROM tbl_posts;
Translating it to Laravel's Eloquent query builder, and using PHP's serialize/unserialize functions, the end result looks like this:将其转换为 Laravel 的 Eloquent 查询构建器,并使用 PHP 的序列化/反序列化函数,最终结果如下所示:
$posts = Post::query()
->select("*")
->selectRaw('(SELECT JSON_ARRAYAGG(JSON_OBJECT(key, value)) FROM tbl_posts_meta WHERE post_id = posts.id) as data');
return $posts->get()->map(function($post){
$meta = [];
foreach($post->data as $dataItem){
foreach($dataItem as $key => $value){
$meta[$key] = unserialize($value);
}
}
$post->data = $meta;
return $post;
});
Thank you for your support and attention!感谢您的支持和关注! :) :)
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