mysql group_concat 问题：ISNULL 替换空值但顺序不正确

Question

使用 GROUP_CONCAT 时，ORDER BY 子句似乎可以正常工作。但是，当使用 ISNULL 或 COALESCE 替换 NULL 值时，尽管实际值是有序的，但 NULL 值都在连接的开头结束（见下文）。如何让我的 NULL 替换值以正确的列顺序排列，以便它们正确地表示被替换的值。

SELECT filledForm.creationDate AS date, GROUP_CONCAT(IFNULL(answers.answer, "NULL") ORDER BY answers.promptID) AS answers
FROM filledForm
JOIN prompts ON filledForm.formID = prompts.formID AND prompts.formID = 100
LEFT JOIN  answers ON prompts.promptID = answers.promptID AND filledForm.filledFormID = answers.filledFormID
GROUP BY filledForm.filledFormID;

使用一些示例数据，这给出：

+---------------------+-----------------------------------------------------------------+
| date                | answers                                                         |
+---------------------+-----------------------------------------------------------------+
| 2023-03-10 14:21:03 | NULL,NULL,NULL,NULL,NULL,bob,2023-03-16,English                 |
| 2023-03-11 02:28:23 | NULL,NULL,NULL,Johnny Depp,60,[email protected],2023-03-24,German |
| 2023-03-11 02:32:41 | NULL,NULL,Tom,22,[email protected],2023-03-29,11:00,English     |
+---------------------+-----------------------------------------------------------------+

代替：

+---------------------+-----------------------------------------------------------------+
| date                | answers                                                         |
+---------------------+-----------------------------------------------------------------+
| 2023-03-10 14:21:03 | bob,NULL,NULL,2023-03-16,NULL,English,NULL,NULL                 |
| 2023-03-11 02:28:23 | Johnny Depp,60,[email protected],2023-03-24,NULL,German,NULL,NULL |
| 2023-03-11 02:32:41 | Tom,22,[email protected],2023-03-29,11:00,EnglishNULL,NULL      |
+---------------------+-----------------------------------------------------------------+

当我在没有按部分分组的情况下运行查询时可以看到：

SELECT answers.answer AS answers
FROM filledForm
JOIN prompts ON filledForm.formID = prompts.formID AND prompts.formID = 100
LEFT JOIN  answers ON prompts.promptID = answers.promptID AND filledForm.filledFormID = answers.filledFormID
ORDER BY filledForm.filledFormID, prompts.promptID;

+-------------------+
| answers           |
+-------------------+
| bob               |
| NULL              |
| NULL              |
| 2023-03-16        |
| NULL              |
| English           |
| NULL              |
| NULL              |
| Johnny Depp       |
| 60                |
| [email protected]   |
| 2023-03-24        |
| NULL              |
| German            |
| NULL              |
| NULL              |
| Tom               |
| 22                |
| [email protected] |
| 2023-03-29        |
| 11:00             |
| English           |
| NULL              |
| NULL              |
+-------------------+

或者通过显示更多的列可能更容易理解：

SELECT filledForm.filledFormID, prompts.promptID, answers.answer AS answers
FROM filledForm
JOIN prompts ON filledForm.formID = prompts.formID AND prompts.formID = 100
LEFT JOIN  answers ON prompts.promptID = answers.promptID AND filledForm.filledFormID = answers.filledFormID
ORDER BY filledForm.filledFormID, prompts.promptID;

+--------------+----------+-------------------+
| filledFormID | promptID | answer            |
+--------------+----------+-------------------+
|           14 |        9 | bob               |
|           14 |       10 | NULL              |
|           14 |       11 | NULL              |
|           14 |       12 | 2023-03-16        |
|           14 |       13 | NULL              |
|           14 |       14 | English           |
|           14 |       15 | NULL              |
|           14 |       16 | NULL              |
|           15 |        9 | Johnny Depp       |
|           15 |       10 | 60                |
|           15 |       11 | [email protected]   |
|           15 |       12 | 2023-03-24        |
|           15 |       13 | NULL              |
|           15 |       14 | German            |
|           15 |       15 | NULL              |
|           15 |       16 | NULL              |
|           16 |        9 | Tom               |
|           16 |       10 | 22                |
|           16 |       11 | [email protected] |
|           16 |       12 | 2023-03-29        |
|           16 |       13 | 11:00             |
|           16 |       14 | English           |
|           16 |       15 | NULL              |
|           16 |       16 | NULL              |
+--------------+----------+-------------------+

这些是完整的表格：

CREATE TABLE prompts (
    promptID INT PRIMARY KEY AUTO_INCREMENT,
    formID INT NOT NULL,
    FOREIGN KEY (formID)
        REFERENCES form (formID)
        ON DELETE CASCADE,        
    hash VARCHAR(16) NOT NULL,
    prompt TEXT(1023) NOT NULL,
    changeDate TIMESTAMP
    UNIQUE(formID, hash)
);


CREATE TABLE filledForm (
    filledFormID INT PRIMARY KEY AUTO_INCREMENT,
    formID INT NOT NULL,
    FOREIGN KEY (formID)
        REFERENCES form (formID)
        ON DELETE CASCADE,
    creationDate TIMESTAMP NOT NULL DEFAULT (UTC_TIMESTAMP),
    INDEX (formID)
);

CREATE TABLE answers (
    id INT PRIMARY KEY AUTO_INCREMENT,
    filledFormID INT NOT NULL, 
    FOREIGN KEY (filledFormID)
        REFERENCES filledForm (filledFormID)
        ON DELETE CASCADE,
    promptID INT NOT NULL,
    FOREIGN KEY (promptID)
        REFERENCES prompts (promptID)
        ON DELETE CASCADE,    
    answer TEXT(99999),
    INDEX (filledFormID)
);

SET SESSION group_concat_max_len = 1000000;

任何帮助将不胜感激。

Answer 1

您正在尝试

ORDER BY answers.promptID

，它位于您

LEFT JOIN

的右侧。

ORDER BY prompts.promptID

代替。

SELECT
    f.creationDate AS date,
    GROUP_CONCAT(IFNULL(a.answer, "NULL") ORDER BY p.promptID) AS answers
FROM filledForm f
JOIN prompts p ON f.formID = p.formID AND p.formID = 100
LEFT JOIN answers a ON p.promptID = a.promptID AND f.filledFormID = a.filledFormID
GROUP BY f.filledFormID;