python之如何在 python/mongodb/elementTree 中将列表项转换为相应的字典

我是 python 的新手。现在是 Python 2.7

我在 elementTree 中处理 xml 并使用 Mongodb。 我要处理的 XML 是“http://www.sec.gov/Archives/edgar/usgaap.rss.xml” 下面是代码:

import os 
import cgi 
import sqlite3 as litefire 
import sys 
sys.stderr = sys.stdout 
from xml.etree import ElementTree 
from pymongo import Connection 
connc2=Connection('localhost',27017) 
db2=connc2['rss'] 
rss=db2.rss 
 
xmlrss=[] 
treexsdr = ElementTree.parse('xbrlrss_all.xml') 
i=0 
k=0 
o=0 
o2=0 
iter = treexsdr.getiterator() 
 
for element in iter: 
    if element.tag: 
        o=i+k 
        xmlname=element.tag 
    if element.keys(): 
        attributedict = dict(element.items()) 
        for name, value in element.items(): 
            krishna=element.items() 
    if element.text: 
        text = element.text 
 
    xmlnamelist={"xmlname":xmlname,"text":text,"ownid":o,"parentid":o2,"xmlattkeys":{k:v for k,v in krishna}} 
 
    xmlrss.append(xmlnamelist) 
 
    if element.getchildren(): 
        o2=o 
        for child in element: 
            k=k+1 
    i=i+1 
 
rss.insert(xmlrss) 

当我应用 krishna=dict(element.items()) 时，我在 IDE 中收到如下错误消息:

Message File Name   Line    Position     
Traceback                
    <module>    D:\test\mongo_rss.py    44       
    insert  C:\Python27\lib\site-packages\pymongo\collection.py 312      
InvalidDocument: key '{http://www.sec.gov/Archives/edgar}file' must not contain '.'  

如果 krishna=element.items()，那么在 mongodb 中我得到:

{ 
  "_id" : ObjectId("4f69bb6e17ea930fd803a958"), 
  "text" : "en-us", 
  "xmlname" : "language", 
  "xmlattkeys" : [["href", "http://www.sec.gov/Archives/edgar/xbrlrss.all.xml"], ["type", "application/rss+xml"], ["rel", "self"]], 
  "parentid" : 2, 
  "ownid" : 16 
} 

但是我想要

{ 
  "_id" : ObjectId("4f69bb6e17ea930fd803a958"), 
  "text" : "en-us", 
  "xmlname" : "language", 
  "xmlattkeys" : {"href":"http://www.sec.gov/Archives/edgar/xbrlrss.all.xml", "type":"application/rss+xml", "rel":"self"}, 
  "parentid" : 2, 
  "ownid" : 16 
} 

请帮助我做到这一点。

请您参考如下方法:

代替

for name, value in element.items(): 
    krishna=element.items() 

做

krishna = dict(element.items()) 

(也许考虑为该变量使用更具描述性的名称。)