Python For循环覆盖字典

Question

我有一个 for 循环将订单从请求返回到在线商店。然后将所需的元素插入到字典中。我的问题是，当我在一个订单中有多个项目时，它会用请求中的最后一个订单覆盖字典。

我可以使用代码中的print("this is an order") 看到订单上有多个商品。

如何增加 dict 中的索引，以便将订单项附加到它而不是覆盖它。

提前致谢。

代码：

   #get specifics on each order with orderID from previous request
    tags2 = ('{http://publicapi.ekmpowershop.com/}ProductCode', '{http://publicapi.ekmpowershop.com/}ProductQuantity',
            '{http://publicapi.ekmpowershop.com/}ProductName', '{http://publicapi.ekmpowershop.com/}OrderDate',
            '{http://publicapi.ekmpowershop.com/}ProductPrice')

    for orderItem in xmlFormatter(ekmSingleOrderRequest(str(list(out.values())[0])), "C:/Users/user/Desktop/test2.xml").iter('{http://publicapi.ekmpowershop.com/}OrderItem'):
        out2 = {}
        #here we can the quantity of order items
        print("this is an order item")
        for child in orderItem:
            count = 0
            if child.tag in tags2:
                out2[child.tag[child.tag.index('}')+1:]] = child.text

Answer 1

使用列表来保存每个项目：

    items = []
    for orderItem in xmlFormatter(ekmSingleOrderRequest(str(list(out.values())[0])), "C:/Users/user/Desktop/test2.xml").iter('{http://publicapi.ekmpowershop.com/}OrderItem'):
        out2 = {}
        #here we can the quantity of order items
        print("this is an order item")
        for child in orderItem:
            if child.tag in tags2:
                out2[child.tag.split('}',1)[1]] = child.text
    items.append(out2)

Answer 2

使用字典重要吗？你可以改用 pandas 数据框吗？

import pandas as pd

df = pd.DataFrame(["A","B"])

for orderItem in xmlFormatter(ekmSingleOrderRequest(str(list(out.values())[0])), "C:/Users/user/Desktop/test2.xml").iter('{http://publicapi.ekmpowershop.com/}OrderItem'):
        out2 = {}
        #here we can the quantity of order items
        print("this is an order item")
        tobeappended = {}
        for child in orderItem:
            count = 0
            if child.tag in tags2:
                tobeappended["A"] = child.tag[child.tag.index('}')+1:]
                tobeappended["B"] = child.text
                df = df.append(tobeappended,ignore_index=True)