Question

当我尝试更新某些表中的信息时，我遇到了一些问题。例如，我有这个表：

class Channel(rdb.Model):
    rdb.metadata(metadata)
    rdb.tablename("channels")

    id = Column("id", Integer, primary_key=True)
    title = Column("title", String(100))
    hash = Column("hash", String(50))
    runtime = Column("runtime", Float)

    items = relationship(MediaItem, secondary="channel_items", order_by=MediaItem.position, backref="channels")

我有这段代码：

def insertXML(channels, strXml):
    channel = Channel()
    session = rdb.Session()
    result = ""

    channel.fromXML(strXml)
    fillChannelTemplate(channel, channels)

    rChannel = session.query(Channel).get(channel.id)
    for chan in channels:
        if rChannel.id == channel.id:
            rChannel.runtime = channel.runtime
            for item in channel.items:
                if item.id == 0:
                    rChannel.items.append(item)

当我执行“rChannel.items.append（item）”时，我收到此错误：

"FlushError: New instance Channel at 0xaf6e48c with identity key
zeppelinlib.channel.ChannelTest.Channel , (152,) conflicts with
persistent instance Channel at 0xac2e8ac"

但是，此指令正在运行“rChannel.runtime = channel.runtime”。

有什么想法吗？

提前致谢

Answer 1

我认为你的代码应该是：

for chan in channels:
    if rChannel.id == channel.id:
        runtime = channel.runtime

或

for chan in channels:
    if rChannel.id == channel.id:
         for item in channel.items:
            if item.id == 0:
                rChannel.items.append(item)

而不是两者兼而有之。在我看来，您将channel.items添加两次到rChannel.items。

更新SQLAlchemy和Python中的表中的记录

1 个答案: