MySQL存储过程在不使用游标的情况下迭代多个值返回错误代码: 1172.结果由多于一行组成

Question

我创建了这个 MySQL 存储过程作为我的问题的解决方案 mentioned in this post on SO .

程序如下:

CREATE DEFINER=`root`@`localhost` PROCEDURE `new_procedure`()
BEGIN
DECLARE n, i INT DEFAULT 0;
DECLARE pid VARCHAR(20);
DROP temporary table if exists tmp;
CREATE temporary table tmp
SELECT 
    t1.patient_id
FROM 
    consultation t1
LEFT JOIN diagnosis t2 
    ON t1.diagnosis_id = t2.diagnosis_id
LEFT JOIN visit t3 
    ON t3.visit_id = t1.visit_id
LEFT JOIN patient t4 
    ON t4.patient_id = t3.patient_id
LEFT JOIN diabetes_assessment t5 
    ON t5.patient_id = t4.patient_id
WHERE 
    t2.diagnosis_name LIKE '%Diabetes%' 
    AND t1.clinic_id = '361'
    AND t3.visit_status="Active"
GROUP BY t1.patient_id ;

set i=1;
SELECT count(*) INTO n FROM tmp;
SELECT patient_id into pid FROM tmp;
while i<=n DO
    set pid = (select patient_id from tmp);
SELECT
    t1.patient_id,
    CONVERT(aes_decrypt(t4.patient_name_en, 'key') USING utf8mb4) as patient_name_en,

    min(t3.date_of_visit) as date_of_visit, 
    t2.diagnosis_name,
    max(ifnull(t5.date_of_assessment, 'N/A')) as date_of_assessment,
    ifnull(t5.assessment_result, 0) as assessment_result 
FROM consultation t1
LEFT JOIN diagnosis t2
    ON t1.diagnosis_id = t2.diagnosis_id
LEFT JOIN visit t3
    ON t3.visit_id = t1.visit_id
LEFT JOIN patient t4
    ON t4.patient_id = t3.patient_id
LEFT JOIN diabetes_assessment t5
    ON t5.patient_id = t4.patient_id
WHERE
   t2.diagnosis_name LIKE '%Diabetes%' AND
   t1.patient_id = pid AND
   t1.clinic_id = '361' AND
   t3.visit_status="Active"
GROUP BY
    t1.patient_id,
    t2.diagnosis_name,
    t3.date_of_visit,
    t4.patient_name_en,
    t5.date_of_assessment,
    t5.assessment_result
ORDER BY t5.date_of_assessment DESC LIMIT 1; 
set i = i + 1;
END WHILE;
END

我会稍微解释一下。以下查询将通过 patent_id 获取 patent_ids group by，因此解决了only full group by 的问题:

DROP temporary table if exists tmp;
CREATE temporary table tmp
SELECT t1.patient_id
FROM consultation t1
LEFT JOIN diagnosis t2 
    ON t1.diagnosis_id = t2.diagnosis_id
LEFT JOIN visit t3 
    ON t3.visit_id = t1.visit_id
LEFT JOIN patient t4 
    ON t4.patient_id = t3.patient_id
LEFT JOIN diabetes_assessment t5 
    ON t5.patient_id = t4.patient_id
WHERE t2.diagnosis_name LIKE '%Diabetes%' AND t1.clinic_id = '361'
AND t3.visit_status="Active"
GROUP BY t1.patient_id ;

然后我将计算这个临时表中有多少行:

set i=1;
SELECT count(*) INTO n FROM tmp;

结果是 2 个患者 ID。

所以我需要迭代每个患者 ID:

select patient_id into pid FROM tmp;

在此查询中:

while i<=n DO
set pid = (select patient_id from tmp);
SELECT
    t1.patient_id,
    CONVERT(aes_decrypt(t4.patient_name_en, 'key') USING utf8mb4) as patient_name_en,
    min(t3.date_of_visit) as date_of_visit, 
    t2.diagnosis_name,
    max(ifnull(t5.date_of_assessment, 'N/A')) as date_of_assessment,
    ifnull(t5.assessment_result, 0) as assessment_result 
FROM consultation t1
LEFT JOIN diagnosis t2
    ON t1.diagnosis_id = t2.diagnosis_id
LEFT JOIN visit t3
    ON t3.visit_id = t1.visit_id
LEFT JOIN patient t4
    ON t4.patient_id = t3.patient_id
LEFT JOIN diabetes_assessment t5
   ON t5.patient_id = t4.patient_id
WHERE
   t2.diagnosis_name LIKE '%Diabetes%' AND
   t1.patient_id = pid AND
   t1.clinic_id = '361' AND
   t3.visit_status="Active"
GROUP BY
    t1.patient_id,
    t2.diagnosis_name,
    t3.date_of_visit,
    t4.patient_name_en,
    t5.date_of_assessment,
    t5.assessment_result
ORDER BY t5.date_of_assessment DESC LIMIT 1; 
set i = i + 1;
END WHILE;

我认为我的问题出在这两行:

select patient_id into pid FROM tmp;

还有

set pid = (select patient_id from tmp);

错误是:

我不想使用游标，正如我们的教授曾经说过的那样，游标非常足智多谋，但却是不好的做法。

Error Code: 1172. Result consisted of more than one row

Answer 1

如果您获得两个不同的患者 ID 并希望循环访问这两个 ID，则需要使用 CURSOR，例如:

DECLARE patient_id_cursor CURSOR FOR SELECT patient_id FROM tmp;

OPEN patient_id_cursor;
FETCH cursor_name INTO pid;
/*
Logic to process `pid`
*/
CLOSE patient_id_cursor;