Question

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))

Answer 1

您转义%%，但然后首先将字符串用作格式化程序：

"...." % table,

返回一个新字符串，其中%%转义百分比被单个%字符替换。 MySQL数据库适配器（ab）也使用% 的字符串格式，因此它将获取该输出并期望能够使用转义的SQL文字填充%s个插槽。正是在那里，SQL语句的'%Y-%m-%d'部分再次被解释为字符串格式并且抛出错误。

解决方案是将倍增加倍：

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,))

或使用str.format()代替，避免双重逃避：

db.execute("SELECT DATE_FORMAT(snapdate,'%%Y-%%m-%%d') AS date, SUM( population ) AS accountpopulation, count( blockid ) AS number_block FROM {0} WHERE blockid = %s GROUP BY snapdate ORDER BY snapdate DESC LIMIT 7".format(table), (blockid,))

此处{0}由表名替换，%%转义未被触及;数据库适配器将使用%s槽填充blockid参数，并返回一个SQL语句，其中%%转义符转换为单个%个字符。

Answer 2

最后，@ Martijn Pieters你完全正确。谢谢你的有用答案。另一个错误来自SUM和COUNT。在处理JSON时，Python有时会以疯狂的方式运行。所以完整的答案是：

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

Python =＆gt; ValueError：不支持的格式字符'Y'（0x59）

2 个答案: