Question

我有以下查询：

SELECT apdr.patient_id,
COUNT(apdr.patient_id)
AS x_ray_result 
FROM apdr
WHERE apdr.type_of_test IN ('x_ray_opinion_1', 'x_ray_opinion_2') 
AND apdr.interpretation LIKE 'kochs' 
GROUP BY apdr.patient_id

它返回解释为'kochs'的患者数量的计数。如何通过输出中的相应patient_id修改此计数以使计数为0。

感谢任何帮助。

我从1开始计数，我需要查询，我也得到0作为计数。

Answer 1

您可以使用case语句返回条件计数，而不是将条件放在where子句中。参见：

SELECT apdr.patient_id, 
    COUNT(CASE WHEN apdr.type_of_test IN ('x_ray_opinion_1', 'x_ray_opinion_2') 
                    AND apdr.interpretation LIKE 'kochs' 
               THEN adpr.patient_id 
          END CASE) AS x_ray_result 
FROM apdr
GROUP BY apdr.patient_id

Answer 2

尝试

SELECT a1.patient_id, 
    (SELECT COUNT(*) FROM apdr a2 
    WHERE a2.interpretation LIKE 'kochs' 
    AND a1.patient_id=a2.patient_id) AS x_ray_result 
FROM apdr a1
WHERE a1.type_of_test IN ('x_ray_opinion_1', 'x_ray_opinion_2')
GROUP BY a1.patient_id

编辑：我忘记了WHERE条款

Answer 3

选择不同的a.patient_id，if（b.x_ray_result为null，0，b.x_ray_result）为来自apdr的x_ray_result左连接（SELECT patient_id，COUNT（patient_id）AS x_ray_result FROM apdr WHERE type_of_test IN（'x_ray_opinion_1'， 'x_ray_opinion_2'）AND解释LIKE'kochs'GROUP BY patient_id）b on a.patient_id = b.patient_id;

MySQL计数与零组合

3 个答案: