MySQL的 - 從表中獲取相關的值

Question

OK我要去嘗試是明確的，因爲我可以，但我忍受 - 它是一個漫長的一週:)

我們有一個結構類似於如下表：

+----+----------+---------+------------------+ 
| id | field_id | user_id |  value  | 
+----+----------+---------+------------------+ 
| 1 | Country |  2 | USA    | 
| 2 | Country |  3 | USA    | 
| 3 | Country |  4 | CA    | 
| 4 | Country |  5 | MX    | 
| 5 | Province |  2 | FL    | 
| 6 | Province |  3 | GA    | 
| 7 | Province |  4 | British Columbia | 
| 8 | Province |  5 | Sonara   | 
| 9 | City  |  2 | Orlando   | 
| 10 | City  |  3 | Brunswick  | 
| 11 | City  |  4 | Vancouver  | 
| 12 | City  |  5 | Nogalas   | 
+----+----------+---------+------------------+` 

，我們需要（希望）查詢返回的所有國家 - 州/省 - 市組合爲動態生成一個JSON文件的目的。

某事的

"SELECT all Provinces and Cities WHERE Country = 'USA'" 

（當然，與我們的數據庫表結構的影響 - 不幸的是不能改變 - 其相當多的複雜

任何的共享「user_id」的值可以安全地假定爲「一起」（即user_id 2的「USA」爲「Country」，「FL」爲「Province」，「Orlando」爲「City」）。

我們試圖創建一個類似於下面的json文件的最終結果。

{ 
"USA": { 
    "Florida": [ 
     {"City": "Orlando"}, 
     {"City": "Palm Beach"} 
    ], 
    "Georgia": [ 
     {"City": "Atlanta"}, 
     {"City": "Brunswick"} 
    ] 
}, 
"Canada": { 
    "Alberta": [ 
     {"City": "Calgary"} 
    ], 
    "Ontario": [ 
     {"City": "Atlanta"}, 
     {"City": "Brunswick"} 
    ] 
} 
} 

Answer 1

有了這個查詢和一些循環的結果，你應該能夠生成您的JSON，只要你想

SQL Fiddle

的MySQL 5.6架構設置：

CREATE TABLE t 
    (`id` int, `field_id` varchar(8), `user_id` int, `value` varchar(16)) 
; 

INSERT INTO t 
    (`id`, `field_id`, `user_id`, `value`) 
VALUES 
    (1, 'Country', 2, 'USA'), 
    (2, 'Country', 3, 'USA'), 
    (3, 'Country', 4, 'CA'), 
    (4, 'Country', 5, 'MX'), 
    (5, 'Province', 2, 'FL'), 
    (6, 'Province', 3, 'GA'), 
    (7, 'Province', 4, 'British Columbia'), 
    (8, 'Province', 5, 'Sonara'), 
    (9, 'City', 2, 'Orlando'), 
    (10, 'City', 3, 'Brunswick'), 
    (11, 'City', 4, 'Vancouver'), 
    (12, 'City', 5, 'Nogalas') 
; 

查詢1：

SELECT 
    tc.`value` as Country, 
    tp.`value` as Province, 
    tcy.`value` as City 
FROM (
     SELECT `value`, user_id FROM t WHERE field_id = "Country" 
    ) as tc 
    LEFT JOIN (
      SELECT `value`, user_id FROM t WHERE field_id = "Province" 
     ) as tp 
     ON tc.user_id = tp.user_id 
    LEFT JOIN (
      SELECT `value`, user_id FROM t WHERE field_id = "City" 
     ) as tcy 
     ON tp.user_id = tcy.user_id 

Results：

|Country|   Province |  City | 
|-------|------------------|-----------| 
| USA |    FL | Orlando | 
| USA |    GA | Brunswick | 
| CA | British Columbia | Vancouver | 
| MX |   Sonara | Nogalas |