如何让查询返回0，而不是空集，如果没有结果

Question

我怎样才能使这个查询返回0值的行，如果有对于每一日期

SELECT COUNT(id) FROM `panel_messages` WHERE `sent_by` = 'root' 
     AND `send_date` IN ("1395-4-25","1395-4-24","1395-4-23","1395-4-22","1395-4-21","1395-4-20","1395-4-19") 
     GROUP BY `send_date` 
     ORDER BY `send_date` DESC 

我期望的结果是没有价值7行是这样的：

| row1 | 

| row2 | 

| row3 | 

| row4 | 

| row5 | 

| row6 | 

| row7 | 

和如果没有结果对于i希望它是0的行中的一个是默认值：

| 2 | 

| 0 | 

| 0 | 

| 2 | 

| 0 | 

| 3 | 

| 1 | 

但是现在我只拿到4行，因为如果没有结果我的查询不返回任何内容：

| 2 | 

| 2 | 

| 3 | 

| 1 | 

SQL小提琴：http://sqlfiddle.com/#!9/a07486/3

Answer 1

试试这个：

SELECT sent_by ,"1395-4-25" as `SEND DATE`,COUNT(*) FROM `panel_messages` WHERE `sent_by` = 'root' AND `send_date` = "1395-4-25" 
union 
SELECT sent_by ,"1395-4-24" as `SEND DATE`,COUNT(*) FROM `panel_messages` WHERE `sent_by` = 'root' AND `send_date` = "1395-4-24" 
union 
SELECT sent_by ,"1395-4-23" as `SEND DATE`,COUNT(*) FROM `panel_messages` WHERE `sent_by` = 'root' AND `send_date` = "1395-4-23" 
ORDER BY `SEND DATE` DESC 

在这种情况下

时的日期没有找到COUNT（*）返回0;但在第一个返回null添加4选择语句，它会返回7行现在它的工作，但它可以更好，如果我发现onother解决方案，我要回到这里

Answer 2

请试一试：

SELECT 
COALESCE(YT.total,t.total) AS cnt 
FROM 
(SELECT 0 AS total) t 
LEFT JOIN 
(
    SELECT 
     COUNT(id) AS total 
    FROM `panel_messages` 
    WHERE `sent_by` = 'root' 
    AND `send_date` IN ("1395-4-25","1395-4-24","1395-4-23","1395-4-22","1395-4-21","1395-4-20","1395-4-19") 
    GROUP BY `send_date` 
    ORDER BY `send_date` DESC 
) YT 
ON 1=1; 

注：

已创建一个虚拟行，其值为0。

后来做这个虚表之间的LEFT JOIN您query

，最后用COALESCE可以实现默认计数0如果你的主查询不返回任何东西。

编辑：

查询：

SELECT 
COALESCE(YT.count,0) AS count 
FROM 
(
    SELECT ADDDATE('1395-01-01', INTERVAL @i:=@i+1 DAY) AS DAY 
    FROM (
    SELECT a.a 
    FROM (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS a 
    CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS b 
    CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS c 
) a 
JOIN (SELECT @i := -1) r1 
) dateTable 
LEFT JOIN 

(
    SELECT 
     send_date, 
     COUNT(id) AS count 
    FROM 
     `panel_messages` 
    WHERE 
     `sent_by` = 'root' 
    AND `send_date` IN (
     "1395-4-25", 
     "1395-4-24", 
     "1395-4-23", 
     "1395-4-20" 
    ) 
    GROUP BY 
     `send_date` 
    ORDER BY 
     `send_date` DESC 
) AS YT 
ON dateTable.DAY = YT.send_date 
WHERE dateTable.DAY IN ('1395-04-25','1395-04-24','1395-04-23','1395-04-20'); 

为了获得零计数不存在，您需要创建一个临时表的日期，所有的时间（下一定范围）居住。

然后在该临时表的日期字段和send_date字段之间进行左连接，可以完成几乎完成的工作。

如果count为NULL，最后您需要使用COALESCE来得到0。

WORKING DEMO

Answer 3

其他答案你想要做什么是不可能的没有工会：

，但你可以尝试一些想象还有 创建包含您的日期

create Table temporary (
     send_date date 
    ); 
    insert INTO temporay("1395-4-25"),("1395-4-24"),("1395-4-23"),("1395-4-22"),("1395-4-21"),("1395-4-20"),("1395-4-19") 

比确实与分辩选择你的桌子，这一个之间加入一个临时表，现在你将有备案日期没有send_by

panel_messages.sent_by | panel_messages.send_date | temporary.send_date 
root       "1395-4-25"    "1395-4-25" 
root       "1395-4-25"    "1395-4-25" 
null       null      "1395-4-24" 
null       null      "1395-4-23" 
root       "1395-4-19"    "1395-4-19" 
. 
. 
. 

现在你在每一天算多少消息，我所做的就是创建一个结果，可以返回你所需要的：

试试这个选择您创建临时表后

SELECT temporary.send_date, count(sender_by) 
from panel_messages RIGTH JOIN temporary ON (temporary.send_date = panel_messages.send_date) 
where 
panel_messages.sent_by like 'root' 
group by temporary.send_date 
ORDER BY send_date DESC;