我目前正在運行此選擇查詢（比原始查詢短）：
SELECT 'website' service, COUNT(*)
FROM website
UNION
SELECT 'vps' service, COUNT(*)
FROM vps
UNION
SELECT 'other' service,
COUNT(*)
FROM other;
我得到以下結果： 結果
--------- ----------
| service | COUNT(*) |
--------- ----------
| website | 2 |
| vps | 1 |
| other | 2 |
--------- ----------
3 rows in set (0.01 sec)
我想將“COUNT(*)”列中列出的所有結果相加。
預期結果：
----------
| COUNT(*) |
----------
| 5 |
----------
1 rows in set (0.00 sec)
我怎樣才能做到這一點？
謝謝！
uj5u.com熱心網友回復：
根據您應該SUM在外部查詢中添加的評論。
嘗試：
SELECT sum(t1.nr_count)
FROM (
SELECT 'website' service, COUNT(*) as nr_count FROM website
UNION
SELECT 'vps' service, COUNT(*) as nr_count FROM vps
UNION
SELECT 'other' service, COUNT(*) as nr_count FROM other
) as t1;
轉載請註明出處，本文鏈接：https://www.uj5u.com/ruanti/369699.html
上一篇：Jest測驗套件無法運行
下一篇：減少選擇中的圈復雜度