3 回答

TA貢獻(xiàn)1757條經(jīng)驗(yàn) 獲得超7個(gè)贊
SELECT t1.ks, t1.[# Tasks], COALESCE(t2.[# Late], 0) AS [# Late]
FROM
(SELECT ks, COUNT(*) AS '# Tasks' FROM Table GROUP BY ks) t1
LEFT JOIN
(SELECT ks, COUNT(*) AS '# Late' FROM Table WHERE Age > Palt GROUP BY ks) t2
ON (t1.ks = t2.ks);

TA貢獻(xiàn)2003條經(jīng)驗(yàn) 獲得超2個(gè)贊
嘗試這樣的事情:
SELECT
*
FROM
(SELECT ks, COUNT(*) AS '# Tasks' FROM Table GROUP BY ks) t1
INNER JOIN
(SELECT ks, COUNT(*) AS '# Late' FROM Table WHERE Age > Palt GROUP BY ks) t2
ON t1.ks = t2.ks

TA貢獻(xiàn)1839條經(jīng)驗(yàn) 獲得超15個(gè)贊
用途UNION:
SELECT ks, COUNT(*) AS '# Tasks' FROM Table GROUP BY ks
UNION
SELECT ks, COUNT(*) AS '# Late' FROM Table WHERE Age > Palt GROUP BY ks
或者,UNION ALL如果您想重復(fù):
SELECT ks, COUNT(*) AS '# Tasks' FROM Table GROUP BY ks
UNION ALL
SELECT ks, COUNT(*) AS '# Late' FROM Table WHERE Age > Palt GROUP BY ks
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