Peewee,MySQL和INSERT IGNORE

时间:2017-01-24 13:31:48

标签: python mysql python-3.x orm peewee

我在一个带有MySQL DB的小Python脚本中使用Peewee作为ORM。

#!/usr/bin/python3
#coding: utf-8

import peewee
from peewee import *

db = MySQLDatabase(**config)
class Foo(peewee.Model):
    bar = peewee.CharField(unique=True, null=False)
    class Meta:
        database = db

try:
    Foo.create_table()
except:
    pass

foo_data = [{'bar':'xyz'},{'bar':'xyz'}]
Foo.insert_many(foo_data).on_conflict(action='IGNORE').execute()

如你所见,我有同样的钥匙。我想第二次使用on_conflict方法(described in the API reference,但仅限SQLite3)忽略它,但是在运行脚本时出现此错误(正常,因为没有为MySQL实现):

peewee.ProgrammingError: (1064, "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 'OR IGNORE INTO `foo` (`bar`) VA' at line 1")

如果删除.on_conflict(action='IGNORE'),MySQL也不喜欢它(重复键)。我怎样才能让peewee插入一个新密钥,或者如果它是一个重复密钥则忽略它?

1 个答案:

答案 0 :(得分:0)

使用db.__str__()。它返回

<peewee.MySQLDatabase object at 0x7f4d6a996198>

如果连接数据库是MySQL和

<peewee.SqliteDatabase object at 0x7fd0f4524198>

如果连接数据库是Sqlite。

所以你可以使用if语句,如:

if 'SqliteDatabase' in db.__str__():
    Foo.insert_many(foo_data).on_conflict(action='IGNORE').execute()
elif 'MySQLDatabase' in db.__str__():
    try:
        Foo.insert_many(foo_data).execute() # or whatever you want with MySQL
    except:
        pass

我认为对于MySQL数据库,你可以这样做:

for data in foo_data:
    for k,v in data.items():
        if (Foo.select(Foo.bar).where(Foo.bar == v).count()) == 0:
            Foo.insert(bar=v).execute()

所以这可以是:

if 'SqliteDatabase' in db.__str__():
    Foo.insert_many(foo_data).on_conflict(action='IGNORE').execute()
elif 'MySQLDatabase' in db.__str__():
    with db.atomic():
        for data in foo_data:
            for k, v in data.items():
                if (Foo.select(Foo.bar).where(Foo.bar == v).count()) == 0:
                    Foo.insert(bar=v).execute()