php - 另一个 table 里面的 table

Question

我想公开计划我的旅行，以便其他人可以加入我。所以，我建立了一个 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     |
+----+---------+---------------+

为了显示，我想在一个查询中查询它们。 我如何查询数据库才能得到这样的东西？(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

首先，您必须使用LEFT JOIN，如下所示:

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

通过使用此功能，您将为每次行程获取一行，并且您可以为每行循环 fellowID 和 twitterName，因为它将是 逗号 分隔列表如下:

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