[英]Mysql query - add data from multiple rows with the same id and place them in a table
[英]MySQL Selecting multiple rows with the same joining ID from the same table
MySQL數據庫示例:
data_name | page_id | data
--------------------------
brand | 83 | brandA
brand | 88 | brandA
brand | 89 | brandA *
type | 89 | typeA *
type | 90 | typeA
type | 98 | typeA
我進入的唯一數據是:
data_name = brand,
data_name = type,
data = brandA,
data = typeA
我需要的是第一個page_id是相同的。 所以在這種情況下,89,89是相同的。 這就是我想要回歸的結果。
我所擁有的幾乎是有效的:
SELECT * FROM `data` WHERE (`data_name` = 'brand' AND `data` = 'brandA') OR (`data_name` = 'type' AND `data` = 'typeA')
我可以用PHP來預測結果,但我更願意在MySQL中找到答案。
有任何想法嗎? 謝謝
只是在這里猜測......
DROP TABLE IF EXISTS my_table;
CREATE TABLE my_table
(data_name VARCHAR(12) NOT NULL
,page_id INT NOT NULL
,data VARCHAR(12) NOT NULL
,PRIMARY KEY(data_name,page_id)
);
INSERT INTO my_table VALUES
('brand',83,'brandA'),
('brand',88,'brandA'),
('brand',89,'brandA'),
('type',89,'typeA'),
('type',90,'typeA'),
('type',98,'typeA');
SELECT x.page_id
FROM my_table x
WHERE (data_name,data) IN(('brand','brandA'),('type','typeA'))
GROUP
BY page_id
HAVING COUNT(*) = 2;
+---------+
| page_id |
+---------+
| 89 |
+---------+
在這種情況下,如果您只想顯示具有品牌和價值的記錄的所有ID
,請嘗試使用此命令,它將顯示您想要的已經具有TYPE和BRAND的每個ID
SELECT A.page_id
FROM data A
INNER JOIN (SELECT * FROM data ) as B ON B.page_id = A.page_id AND B.data_name != A.data_name
GROUP BY A.page_id
這應該有效,將表連接到自身:
SELECT
d1.page_id
FROM
(SELECT
pade_id
FROM
`data`
WHERE
data_name = 'type') d1
INNER JOIN
(SELECT
pade_id
FROM
`data`
WHERE
data_name = 'brand') d2
ON d1.page_id = d2.page_id;
應該管用:
SELECT orig.page_id, second.page_id as s_page_id
FROM `data` as orig, (SELECT * FROM `data` WHERE page_id = orig.page_id AND data_name = 'brand') AS second
WHERE orig.data_name = 'type';
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