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Update a record in a table in SQLAlchemy and Python

开发者 https://www.devze.com 2023-01-15 07:07 出处:网络
I have some problems when I try to update information in some tables. For example, I have this table:

I have some problems when I try to update information in some tables. For example, I have this table:

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")

And I have this code:

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 ch开发者_运维问答annels:
        if rChannel.id == channel.id:
            rChannel.runtime = channel.runtime
            for item in channel.items:
                if item.id == 0:
                    rChannel.items.append(item)

When I do "rChannel.items.append(item)", I got this error:

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

However, this instruction is working "rChannel.runtime = channel.runtime".

Any ideas?

Thanks in advance


I think your code should be either:

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

or

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

But not both. It appears to me that you're adding channel.items twice to rChannel.items.

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