Question

代码应该从CSV中获取由另一种方法生成的字典 - 并生成所有必需的客户和订单实体。

   def loadOrderData(fileName,session):
    print("Orders loading")
    liDict = Loader.csv2liDict(fileName, {1: "date", 16: "customerName", 22: "customerPostcode", 23: "salesOrderNo",
                                   25: "worksOrderNo", 35: "productCode", 38: "width", 39: "length",
                                   53: "quantity"})
    for i in liDict:
        customerId = -1
        for j in session.query(Customer.id). \
                filter(Customer.name == i["customerName"]). \
                filter(Customer.postcodeDistrict == i["customerPostcode"].split(" ")[0]):
            customerId = j
        if customerId == -1:
            newCustomer = Customer(name=i["customerName"], postcodeDistrict=i["customerPostcode"].split(" ")[0])
            session.add(newCustomer)
            session.commit()
            customerId = newCustomer.id
        print("CUSTOMER ID : ",customerId)
        newOrder = Order(date=str2date(i["date"]), customerId=customerId, salesOrderNo=i["salesOrderNo"],
                         worksOrderNo=i["worksOrderNo"], productCode=i["productCode"], width=int(i["width"]),
                         length=int(i["length"]), quantity=int(i["quantity"]))
        session.add(newOrder)
        session.commit()

我一直收到以下错误：

 sqlalchemy.exc.InterfaceError: (raised as a result of Query-invoked autoflush; consider using a session.no_autoflush block if this flush is occurring prematurely) (sqlite3.InterfaceError) Error binding parameter 1 - probably unsupported type. [SQL: 'INSERT INTO "order" (date, "customerId", "worksOrderNo", "salesOrderNo", "productCode", width, length, quantity, assigned) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?)'] [parameters: ('2016-10-26 00:00:00.000000', (1,), '', 'S/O269155', 'BKT1', 724, 1769, 0, None)]

基本上，我推断它是由于customerId等于（1，）而不是int。但是，我不明白为什么会发生这种情况以及如何解决这个问题。建议，请。

Answer 1

可以看到here，通过用逗号分隔项来创建元组(1,)。可以使用尾随逗号创建一个项目元组，这通常会引起混淆。

customerId = 1,

尝试调试这些情况时，最好将注意力集中在项目的创建位置。在你的例子的情况下，我建议从csv解析开始，但由于我无法运行此代码，我不能再给出具体的建议。

更新

根据roganjosh的评论，该项目由查询返回。它设置的查询接口使得可以返回多个结果，因此可能返回元组。您应该调查Query.one()或Query.one_or_none。

Answer 2

我自己找到了答案：我复制了循环查询并将其放在if customerId == -1语句的末尾，如下所示：

                for j in session.query(Customer.id). \
                filter(Customer.name == i["customerName"]). \
                filter(Customer.postcodeDistrict == i["customerPostcode"].split(" ")[0]):
            customerId = j[0]
        if customerId == -1:
            newCustomer = Customer(name=i["customerName"], postcodeDistrict=i["customerPostcode"].split(" ")[0])
            session.add(newCustomer)
            session.commit()
            for j in session.query(Customer.id). \
                    filter(Customer.name == i["customerName"]). \
                    filter(Customer.postcodeDistrict == i["customerPostcode"].split(" ")[0]):
                customerId = j[0]

SQLAlchemy：object.id在括号中返回int而不是int

2 个答案: