Table 在另一个里面

Question

我想公开计划我的旅行，以便其他人可以加入我的行列。所以，我已经建立了一个 PHP 网站。

我有这个表：

trips:
+----+---------+------------+------------+-------------------------+
| id | title   | date_start | date_end   | marker_adress           |
+----+---------+------------+------------+-------------------------+
|  1 | Berlin  | 2015-07-10 | 2015-07-11 | Potsdamer Platz, Berlin |
|  2 | Hamburg | 2015-07-16 | 2015-07-18 | Jungfernstieg, Hamburg  |
+----+---------+------------+------------+-------------------------+

fellows:
+----+---------+---------------+
| id | trip_id | twittername   |
+----+---------+---------------+
|  1 |       1 | prtyengopls   |
|  2 |       1 | itobi_yt      |
|  3 |       1 | jessisadancer |
|  4 |       2 | jessisadancer |
|  5 |       2 | woelfch3n     |
+----+---------+---------------+

为了展示，我想一次查询。 我怎样才能查询数据库，所以我有这样的东西？（我知道，它是 JSON 但它很好地显示了结构。）

{
    "id": 1,
    "date_start": "2015-07-10",
    "date_end": "2015-07-11",
    "marker_adress": "Potsdamer Platz, Berlin",
    "fellows": [
        {
            "id": 1,
            "twittername": "prtyengopls"
        },
        {
            "id": 2,
            "twittername": "itobi_yt"
        },
        {
            "id": 3,
            "twittername": "jessisadancer"
        }
    ]
}

Answer 1

我认为这不可能像您在 SQL 中预期的那样。只是为了回答您的问题并向您展示问题：您必须使用联接（左联接、内联接或其他……取决于您的数据库结构；如果可能的话，我总是更喜欢左联接）来仅在一个查询中获取所有信息：

SELECT * FROM trips t LEFT JOIN fellows f on t.id = f.trip_id WHERE t.id = 1;

但是你总是会得到每一行的行程信息，为此你必须事后处理它。你永远无法 select 这样的嵌套结构，你将永远得到一个扁平的。所以我建议将它分成两个查询，如下所示：

SELECT * FROM trips WHERE id = 1;
SELECT * FROM fellows WHERE trip_id = 1;

您之后也必须处理这些信息，但您只是 select 来自数据库的所需信息。

希望对您有所帮助。

Answer 2

首先你必须像这样使用 LEFT JOIN:

SELECT 
  t.id AS tripID, 
  t.title AS title, 
  t.date_start AS dateStart, 
  t.date_end AS dateEnd, 
  t.marker_address AS markerAddress,
  GROUP_CONCAT(DISTINCT CAST(f.id AS CHAR)) AS fellowID,
  GROUP_CONCAT(DISTINCT CAST(f.twittername AS CHAR)) AS twitterName
FROM trips t LEFT JOIN fellows f ON t.id = f.trip_id
GROUP BY t.id

通过使用它，您将为每次旅行获得一行，并且您可以为每一行遍历 fellowID 和 twitterName，因为它将是 comma 分隔列表，如下所示:

fellowID: 1,2,3
twitterName: prtyengopls,itobi_yt,jessisadancer

Edit 1: I got a new column to trips called checked which is a boolean. Could you update your query, so only trips that have this boolean toggled on are displayed?

SELECT 
  t.id AS tripID, 
  t.title AS title, 
  t.date_start AS dateStart, 
  t.date_end AS dateEnd, 
  t.marker_address AS markerAddress,
  GROUP_CONCAT(DISTINCT CAST(f.id AS CHAR)) AS fellowID,
  GROUP_CONCAT(DISTINCT CAST(f.twittername AS CHAR)) AS twitterName
FROM trips t INNER JOIN fellows f 
  ON t.id = f.trip_id AND t.checked = 1
GROUP BY t.id

Table 在另一个里面

Table inside of another

php

mysql

select