小编OlZ*_*OlZ的帖子

我不理解Y的ValueError.我用%转义...

table = town+"_history"
db.execute("SELECT DATE_FORMAT(snapdate,'%%Y-%%m-%%d') AS date, SUM( population ) AS accountpopulation, count( blockid ) AS number_block FROM %s WHERE blockid =%%s GROUP BY snapdate ORDER BY snapdate DESC LIMIT 7" % table, (blockid))

SELECT t1.id, t1.name, t1.population, CAST(SUM(t2.town_1) AS CHAR) AS town, CAST(SUM(CASE WHEN t2.id LIKE 23 THEN 1 ELSE 0 END) AS CHAR) AS population FROM area1 t1 LEFT JOIN area2 t2 ON t1.id = t2.id WHERE t1.id like 23

通常,如果两个表中都存在id = 23,它将匹配.但事实并非如此(不在t1表中),因此请求返回NULL值.
我怎么做才能没有结果？

我有3个查询工作完美

SELECT snapdate, COUNT( DISTINCT uid) AS t1
FROM table where tid =1
GROUP BY snapdate
ORDER BY snapdate DESC LIMIT 7

查询#1:

snapdate        t1
----------      --
2013-07-02      10
2013-07-01      20
2013-06-30      60
2013-06-29      80
2013-06-28      3
2013-06-27      22
2013-06-26      93

查询#2:

SELECT snapdate, COUNT( DISTINCT uid) AS t2
FROM table where tid =2
GROUP BY snapdate
ORDER BY snapdate DESC LIMIT 7

snapdate        t2
----------      --
2013-07-02      35
2013-07-01      52
2013-06-30      69
2013-06-29      75
2013-06-28      91
2013-06-27      97
2013-06-26 …