2010-09-13 128 views
2

我創建了一個表(MySQL 5.1中)SQLAlchemy的Select語句 - SQL語法錯誤

from sqlalchemy import * 

def get(): 
    db = create_engine('mysql://user:[email protected]/database') 
    db.echo = True 
    metadata = MetaData(db) 

    feeds = Table('feeds', metadata, 
      Column('id', Integer, primary_key=True), 
      Column('title', String(100)), 
      Column('link', String(255)), 
      Column('description', String(255)), 
    ) 

    entries = Table('entries', metadata, 
      Column('id', Integer, primary_key=True), 
      Column('fid', Integer), 
      Column('url', String(255)), 
      Column('title', String(255)), 
      Column('content', String(5000)), 
      Column('date', DateTime), 
    ) 
    feeds.create() 
    entries.create() 

但是當我嘗試進行查詢:

from sqlalchemy import * 
db = create_engine('mysql://user:[email protected]/database') 
metadata = MetaData(db) 
feeds = Table('feeds', metadata) 
s = feeds.select() 
result = db.execute(s) 

我得到的結果的誤差= DB .execute(S)線指示如下:

sqlalchemy.exc.ProgrammingError: (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 'FROM feeds' at line 2") 'SELECT \nFROM feeds'() 

我顯然新SQLAlchemy的,我不知道我做錯了,儘管HAV在網絡上搜索每一個教程並改變了這一百萬次。任何幫助?

回答

0

您的feeds.select()調用中可能缺少某些東西,我會再看看此函數的API文檔。