如何将列表项转换为python/mongodb/elementTree中的相应dict

Question

我是python的新手.现在它的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中得到的错误信息如下:

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
}

请帮助我这样做.

Answer 1

代替

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

做

krishna = dict(element.items())

(也许可以考虑为这个变量使用更具描述性的名称.)