MySQL的JSON LEFT JOIN

Question

TABLE streams; 

+----+----------+-------------------------------------------------------------+-------------+-------------+ 
| id | name  | stream              | server  | bouquet  | 
+----+----------+-------------------------------------------------------------+-------------+-------------+ 
| 1 | HOME  | ["http://ur11.com/", "http://ur12.com/"]     | ["1", "3"] | ["3","2"] | 
+----+----------+-------------------------------------------------------------+-------------+-------------+ 

TABLE servers; 

+----+-------------------+ 
| id | server   | 
+----+-------------------+ 
| 1 | Main    | 
| 2 | Server #1   | 
| 3 | Server #2   | 
| 4 | Server #3   | 
| 5 | Server #4   | 
+----+-------------------+ 

我需要運行LEFT JOIN表流和使用servers.server的streams.server值第一個索引檢索：

比如我需要左連接來得到這樣的結果：

+----+----------+-------------------------------------------------------------+-------------+-------------+ 
| id | name  | stream              | server  | bouquet  | 
+----+----------+-------------------------------------------------------------+-------------+-------------+ 
| 1 | HOME  | ["http://ur11.com/", "http://ur12.com/"]     | Main  | ["3","2"] | 
+----+----------+-------------------------------------------------------------+-------------+-------------+ 

所以關鍵是要得到streams.server女巫第一指數爲1，然後LEFT JOIN上servers.id得到servers.server名女巫是在這種情況下，主要

我mysql安裝5.7版本witc h包含JSON支持，但我沒有弄清楚如何在JSON字段類型上使用LEFT JOIN？

Answer 1

我想你想：

select s.*, se.server 
from streams s join 
    servers se 
    on se.id = json_extract(s.server, '$[0]'); 

根據你安裝了MySQL的確切版本，你也可以使用s.server->'$[0]'。

Answer 2

例如：

SELECT * 
FROM foo 
CROSS JOIN LATERAL json_array_elements (foo.bars :: json) foo_bars 
LEFT OUTER JOIN bar ON (foo_bars ->> 'id') :: BIGINT = bar.ID; 

Link

Answer 3

我得到了它workijng使用上述解決方案：

select s.*, se.server 
from streams s left join 
servers se 
on se.id = CAST(json_extract(s.server, '$[0]') AS UNSIGNED); 

感謝。