Question

我的问题是如何连续获取所有数据（以可打印格式，而不仅仅是对象），用户可以在其中指定所需数据列的名称。这是我目前的代码：

#!/usr/bin/python3
from sqlalchemy import create_engine, MetaData, Table, Column, String, Integer
from sqlalchemy.orm import mapper, sessionmaker
from sqlalchemy import create_engine
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import sessionmaker

class SQL_Engine:

    def __init__(self, dbpath, username, password, dbtype='sqlite', tableName='content_blocks', connector=None):
            self.dbpath = dbpath
            self.username = username
            self.password = password
            self.dbtype = dbtype
            self.tableName = tableName
            self.connector = connector

    def Create_Session(self):
            if self.connector == None:
                    engine = create_engine(self.dbtype+"://"+self.username+":"+self.password+"@"+self.dbpath)
            else:
                    engine = create_engine(self.dbtype+"+"+self.connector+"://"+self.username+":"+self.password+"@"+self.dbpath)
            Base = declarative_base(engine)
            class ContentBlocks(Base):
                    __tablename__ = self.tableName
                    __table_args__ = {'autoload':True}

            metadata = Base.metadata
            Session = sessionmaker(bind=engine)
            session = Session()
            return session, ContentBlocks, Base

    def ViewTable(self):
            session, ContentBlocks, Base = self.Create_Session()
            request = session.query(ContentBlocks).all()
            columns = ContentBlocks.__table__.columns.keys()
            table = " ".join(columns)
            for entry in request:
                    for column in columns:
                            #print entry.column; can't do this as of yet

我在Create_Session()中看到的初始尝试是检索所有列名并将它们作为变量传递给entry对象，但这只会导致错误：

Traceback (most recent call last):
   File "SQLDeduplication-copy.py", line 131, in <module>
      no.ViewTable()
   File "SQLDeduplication-copy.py", line 64, in ViewTable
      print(entry.column)
   AttributeError: 'ContentBlocks' object has no attribute 'column'

因此，我需要将python解释column作为指针，而不是字面意思。这就是我碰壁的地方，因为我见过的所有其他StackOverflow答案都包括使用列名。

那就是说，如果有人有更好的想法，我对此完全开放。如果有任何错误，请更正我的代码

Answer 1

IljaEverilä有正确的解决方案。这是：

def ViewTable(self):
    session, ContentBlocks, Base = self.Create_Session()
    request = session.query(ContentBlocks).all()
    columns = ContentBlocks.__table__.columns.keys()
    num = 1
    for entry in request:
        print(num, '\', end=' ')
        for column in columns:
            print(column, ':', getattr(entry, column), '|',  end=' ')

在不知道sqlalchemy中的列名的情况下获取整行的数据

1 个答案: