Question

我有一张包含斧头的桌子，是的，我还有另一张包含斧头的桌子，以及其他一些东西。

我正在尝试选择总斧数，斧头插入b中，斧头未插入b中。

我有

SELECT
    COUNT(a.ax) AS p,
    COUNT(a.ax) AS q,
FROM 
    a
LEFT JOIN b ON a.ay = b.ay
WHERE b.ay is NULL;

这给我一张桌子 - 但两个都是2.我不确定如何在保持左连接的同时分离条件。

感谢任何建议 - 谢谢。

Answer 1

看来你需要这样的东西：

SELECT
    COUNT(*) AS total,
    COUNT(r.hnum) AS inserted
FROM 
    h
LEFT JOIN r ON h.hnum = r.hnum;

或者你可以使用条件聚合：

SELECT
    COUNT(*) AS total,
    SUM(CASE WHEN r.hnum IS NULL THEN 1 ELSE 0 END) As noninserted,
    SUM(CASE WHEN r.hnum IS NULL THEN 0 ELSE 1 END) As inserted
FROM 
    h
LEFT JOIN r ON h.hnum = r.hnum;

此处有SQLFiddle Demo条件聚合。

处理r中的重复项：

SELECT
    COUNT(*) AS total,
    SUM(CASE WHEN r.hnum IS NULL THEN 1 ELSE 0 END) As noninserted,
    SUM(CASE WHEN r.hnum IS NULL THEN 0 ELSE 1 END) As inserted
FROM 
    h
LEFT JOIN (SELECT DISTINCT hnum FROM r) r ON h.hnum = r.hnum;

这里的SQLFiddle Demo用于带有重复项的条件聚合。

带有计算字段的

JOIN和COUNT (DISTINCT)：

SELECT result.total As "Total", 
       result.inserted as "Inserted", 
       result.total - result.inserted as "Noninserted"
FROM

( SELECT
    COUNT(DISTINCT h.hnum) AS total,
    COUNT(DISTINCT r.hnum) AS inserted
FROM 
    h
LEFT JOIN r ON h.hnum = r.hnum ) result;

以及此Fiddle

sql同列不同的条件

1 个答案: