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私はpython mysqlコネクタを初めて使用し、p_id = 0のuser_idを見つけるクエリの結果を取得しようとしています製品テーブルに移動し、その都市で利用可能な製品の数を見つけます

import mysql.connector
con = mysql.connector.connect(user='user', password = 'pass', host = 'blah.com')

cursor1 = con.cursor(buffered = True)

query = ("SELECT l.user_id, count(u.prod_id)"
        "FROM db1.users as l INNER JOIN db2.product as u "
        "ON l.u_city = u.p_city"
        "WHERE l.p_id =0 GROUP BY l.user_id limit 10;" )
cursor1.execute(query)

クエリはmysqlを実行していますが、python mysqlコネクタから次のエラーが発生しています

C:\Python27\python.exe C:/Python27/Lib/site-packages/mysql/connector/update_campus_user_profile_suset.py
Traceback (most recent call last):

File "C:/Python27/Lib/site-packages/mysql/connector/update_campus_user_profile_suset.py", line 12, in <module>
cursor1.execute(camp_zip)

File "C:\Python27\lib\site-packages\mysql\connector\cursor.py", line 491, in execute
self._handle_result(self._connection.cmd_query(stmt))

File "C:\Python27\lib\site-packages\mysql\connector\connection.py", line 683, in cmd_query
statement))

File "C:\Python27\lib\site-packages\mysql\connector\connection.py", line 601, in _handle_result
raise errors.get_exception(packet)
mysql.connector.errors.ProgrammingError: 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'l.campus_id <2 GROUP BY l.user_id' at line 1

Process finished with exit code 1
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2 に答える 2

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query ='''SELECT l.user_id, count(u.prod_id) FROM db1.users as l INNER JOIN db2.product as u ON l.u_city = u.p_city WHERE l.p_id =0 GROUP BY l.user_id limit 10'''
  • 常にクエリを ('''stmt''') で変数に初期化して実行することをお勧めします
  • 使用する必要はありません;
于 2014-06-12T02:27:13.787 に答える