SQL查询根据COUNT结果返回Yes或No的实现问题咨询

你的SQL无法运行是语法错误：CASE表达式前面多了一个未闭合的左括号，导致SQL解析器无法正常识别语法结构。

方案1：直接修正原有写法 #

删除CASE前多余的左括号即可正常运行，修正后的完整SQL如下：

SELECT 
    a.*, 
    own1.naam AS resposible_btw, 
    own2.naam AS resposible_client, 
    CASE 
        WHEN (SELECT COUNT(planning_clientid) FROM tbl_planning WHERE planning_afgehandeld = 0 AND a.client_id = planning_clientid) > 0 
        THEN 'Yes' 
        ELSE 'No' 
    END AS timesonplanning 
FROM tbl_clients a 
LEFT JOIN gebruikers own1 ON own1.id = a.client_ownerob 
LEFT JOIN gebruikers own2 ON own2.id = a.client_owner

方案2：更高效的EXISTS写法（推荐） #

COUNT统计需要扫描所有符合条件的行才能返回结果，如果你仅需要判断是否存在符合条件的记录，用EXISTS性能更高，尤其在tbl_planning表数据量较大时优势非常明显：

SELECT 
    a.*, 
    own1.naam AS resposible_btw, 
    own2.naam AS resposible_client, 
    CASE 
        WHEN EXISTS (SELECT 1 FROM tbl_planning WHERE planning_afgehandeld = 0 AND a.client_id = planning_clientid)
        THEN 'Yes' 
        ELSE 'No' 
    END AS timesonplanning 
FROM tbl_clients a 
LEFT JOIN gebruikers own1 ON own1.id = a.client_ownerob 
LEFT JOIN gebruikers own2 ON own2.id = a.client_owner

两种方案最终返回的timesonplanning字段都只会输出Yes/No，完全符合你的需求。

内容的提问来源于stack exchange，提问作者Martin15789