MySQL從備用組中選擇

查詢時遇到一些麻煩。我試圖創建從表中選擇deposit_type_language_id = 3或如果行不存在1（作爲後備）與deposit_type_id分組。MySQL從備用組中選擇

+-----------------+--------------------------+-------------------+ 
    | deposit_type_id | deposit_type_language_id | deposit_type_name | 
    +-----------------+--------------------------+-------------------+ 
    |    1 |      1 | jewellery   | 
    |    1 |      2 | bijuterii   | 
    |    1 |      3 | šperky   | 
    |    2 |      1 | equipment   | 
    |    2 |      2 | tehnica   | 
    +-----------------+--------------------------+-------------------+

我試圖用下面的查詢：

SELECT t1.* from deposit_type AS t1 
     INNER JOIN 
      (SELECT deposit_type_id, min(deposit_type_language_id) from 
        deposit_type group by deposit_type_id) AS t2 
     ON t1.deposit_type_id=t2.deposit_type_id group by deposit_type_id;

輸出是：

+-----------------+--------------------------+-------------------+ 
| deposit_type_id | deposit_type_language_id | deposit_type_name | 
+-----------------+--------------------------+-------------------+ 
|    1 |      1 | jewellery   | 
|    2 |      1 | equipment   | 
+-----------------+--------------------------+-------------------+

但我試圖實現類似如下：

+-----------------+--------------------------+-------------------+ 
| deposit_type_id | deposit_type_language_id | deposit_type_name | 
+-----------------+--------------------------+-------------------+ 
|    1 |      3 | šperky   | 
|    2 |      1 | equipment   | 
+-----------------+--------------------------+-------------------+

在哪裏我的錯？

表結構：

deposit_type | CREATE TABLE `deposit_type` (
    `deposit_type_id` smallint(5) unsigned NOT NULL, 
    `deposit_type_language_id` smallint(5) unsigned NOT NULL, 
    `deposit_type_name` varchar(96) NOT NULL, 
    PRIMARY KEY (`deposit_type_id`,`deposit_type_language_id`), 
    KEY `deposit_type_language_id` (`deposit_type_language_id`), 
    CONSTRAINT `deposit_type_ibfk_1` FOREIGN KEY (`deposit_type_language_id`) REFERENCES `language` (`language_id`) 
) ENGINE=InnoDB DEFAULT CHARSET=utf8 |

來源

2015-10-09 vit

仇敵會討厭......

SELECT x.deposit_type_id 
    , COALESCE(y.deposit_type_language_id,x.deposit_type_language_id) deposit_type_language_id 
    , COALESCE(y.deposit_type_name,x.deposit_type_name) deposit_type_name 
    FROM deposit_type x 
    LEFT 
    JOIN deposit_type y 
    ON y.deposit_type_id = x.deposit_type_id 
    AND y.deposit_type_language_id = 3 
WHERE x.deposit_type_language_id = 1;

來源

2015-10-09 15:30:39 Strawberry

@Strawberrry，嗨，哥們你的答案似乎是正確的。我正在學習你的答案，謝謝！ – vit

如果你想用兩個字解釋你的想法，請。 – vit

@vit如果存在有效的連接，則左[outer]連接將返回值，否則返回NULL。然後COALESCE簡單地忽略NULL值。 – Strawberry

@CFreitas，我恨這個男人誰投下來你。你回答不正確，但給了我一個想法。我們應該增加一個比較，見下文。

SELECT t1.* from deposit_type AS t1 
     INNER JOIN 
       (SELECT deposit_type_id, max(deposit_type_language_id) as dtl FROM deposit_type 
        WHERE deposit_type_language_id=3 or deposit_type_language_id=1 group by deposit_type_id) AS t2 
       ON t1.deposit_type_id=t2.deposit_type_id and t1.deposit_type_language_id=t2.dtl group by deposit_type_id;

輸出是：

+-----------------+--------------------------+-------------------+ 
| deposit_type_id | deposit_type_language_id | deposit_type_name | 
+-----------------+--------------------------+-------------------+ 
|    1 |      3 | šperky   | 
|    2 |      1 | equipment   | 
+-----------------+--------------------------+-------------------+

來源

2015-10-09 14:53:36 vit

你可以最多投票我呢:) – CFreitas

MySQL從備用組中選擇

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