如果列--MySQL中没有结果，则获取0值

我有以下疑问

SELECT COUNT( iContactId ) AS Users, DATE( dCreatedAt ) AS ActivityDate 
FROM contacts WHERE iAppId =".$iAppId." 
AND DATE(dCreatedAt) IN (CURRENT_DATE(), CURRENT_DATE()-1 ) 
GROUP BY ActivityDate 

这样我就明白了

Users |ActivityDate
1     |2014-09-19

如果没有匹配行，则我需要0

e、 g

我该怎么做。

查询 你可以用工会

SELECT COUNT( iContactId ) AS Users, CURRENT_DATE() AS ActivityDate
FROM contacts WHERE iAppId =".$iAppId." 
AND DATE(dCreatedAt) = CURRENT_DATE()
UNION
SELECT COUNT( iContactId ) AS Users, CURRENT_DATE() - 1 AS ActivityDate 
FROM contacts WHERE iAppId =".$iAppId." 
AND DATE(dCreatedAt) = CURRENT_DATE() - 1

查询

---

在大多数答案中，这是用于日期减法：

SELECT CURRENT_DATE() - 1
-> 20140919

这将导致隐式类型转换，并且日期将转换为int，请改用此选项：

SELECT CURRENT_DATE - INTERVAL 1 DAY
-> 2014-09-19

我的解决方案

SELECT
    d.ActivityDate,
    COUNT(c.iContactId) Users
FROM
    (
    SELECT CURRENT_DATE - INTERVAL 1 DAY ActivityDate FROM DUAL
    UNION
    SELECT CURRENT_DATE FROM DUAL
    ) d
LEFT JOIN
    contacts c
    ON DATE(c.dCreatedAt) = d.ActivityDate
    AND c.iAppId = " . $iAppId . "
GROUP BY d.ActivityDate

---

很抱歉，忘记删除计数（）我需要计数和日期。@pln需要您解释。我在匹配的日期行中获得20140920作为日期格式

SELECT CURRENT_DATE() - 1
-> 20140919

SELECT CURRENT_DATE - INTERVAL 1 DAY
-> 2014-09-19

SELECT
    d.ActivityDate,
    COUNT(c.iContactId) Users
FROM
    (
    SELECT CURRENT_DATE - INTERVAL 1 DAY ActivityDate FROM DUAL
    UNION
    SELECT CURRENT_DATE FROM DUAL
    ) d
LEFT JOIN
    contacts c
    ON DATE(c.dCreatedAt) = d.ActivityDate
    AND c.iAppId = " . $iAppId . "
GROUP BY d.ActivityDate