我在 Python 中执行一些 SQL 时遇到问题,尽管类似的 SQL 在mysql
命令行。
该表如下所示:
mysql> SELECT * FROM foo;
+-------+-----+
| fooid | bar |
+-------+-----+
| 1 | A |
| 2 | B |
| 3 | C |
| 4 | D |
+-------+-----+
4 rows in set (0.00 sec)
我可以从 mysql 命令行执行以下 SQL 查询,没有问题:
mysql> SELECT fooid FROM foo WHERE bar IN ('A','C');
SELECT fooid FROM foo WHERE bar IN ('A','C');
+-------+
| fooid |
+-------+
| 1 |
| 3 |
+-------+
2 rows in set (0.00 sec)
然而,当我尝试在 Python 中执行相同操作时,我没有得到任何行,而我期望得到 2 行:
import MySQLdb
import config
connection=MySQLdb.connect(
host=config.HOST,user=config.USER,passwd=config.PASS,db='test')
cursor=connection.cursor()
sql='SELECT fooid FROM foo WHERE bar IN %s'
args=[['A','C']]
cursor.execute(sql,args)
data=cursor.fetchall()
print(data)
# ()
所以问题是:应该如何修改python代码来选择那些fooid
s where bar
is in ('A','C')
?
顺便说一句,我注意到如果我交换角色bar
and fooid
,我可以获得选择那些的代码bar
s where fooid
is in (1,3)
成功地。我不明白为什么一个这样的查询(下面)有效,而另一个查询(上面)却不起作用。
sql='SELECT bar FROM foo WHERE fooid IN %s'
args=[[1,3]]
cursor.execute(sql,args)
data=cursor.fetchall()
print(data)
# (('A',), ('C',))
绝对清楚的是,这就是foo
表已创建:
mysql> DROP TABLE IF EXISTS foo;
Query OK, 0 rows affected (0.00 sec)
mysql> CREATE TABLE `foo` (
`fooid` int(11) NOT NULL AUTO_INCREMENT,
`bar` varchar(10) NOT NULL,
PRIMARY KEY (`fooid`));
Query OK, 0 rows affected (0.01 sec)
mysql> INSERT into foo (bar) values ('A'),('B'),('C'),('D');
Query OK, 4 rows affected (0.00 sec)
Records: 4 Duplicates: 0 Warnings: 0
Edit:当我启用一般查询日志时mysqld -l /tmp/myquery.log
I see
mysqld, Version: 5.1.37-1ubuntu5.5-log ((Ubuntu)). started with:
Tcp port: 3306 Unix socket: /var/run/mysqld/mysqld.sock
Time Id Command Argument
110101 11:45:41 1 Connect unutbu@localhost on test
1 Query set autocommit=0
1 Query SELECT fooid FROM foo WHERE bar IN ("'A'", "'C'")
1 Query SELECT bar FROM foo WHERE fooid IN ('1', '3')
1 Quit
事实上,看起来引用太多了A
and C
.
感谢@Amber 的评论,我更好地理解出了什么问题。 MySQLdb 转换参数化参数['A','C']
to ("'A'","'C'")
.
有没有办法使用参数化查询IN
SQL语法?或者必须手动构造 SQL 字符串?