Question

我需要一些帮助，我正在努力做到： 修改每个键的单个字典值（在这种情况下为param），一次只为一个参数，每次迭代的值为self.fooz。

喜欢这个

这样的网址就像：somesite.com?id=6&name=Bill一样 somesite.com?id=<self.fooz>&name=Bill（迭代数量个人愚蠢）然后，somesite.com?id=6&name=<self.fooz> （迭代个别傻瓜的数量）

最后，生成full_param_vector和full_param值，如下所述

有人可以帮帮我吗？

我做过： 1）通过self.path_object引入一组原始路径 2）在?之后解析路径以获取所有原始参数化key/values（通过parse_after）

我写下了一些我想要完成的伪代码：

if self.path_object is not None:
    dictpath = {}
    for path in self.path_object:
        #path.pathToScan - returns a full url e.g. somesite.com?id=6&name=Bill
        #parse_after returns a string with parameters only, like: {u'id': [u'2'], u'name': [u'Dog']}
        parse_after = urlparse.parse_qs(path.pathToScan[path.pathToScan.find('?') + 1:], keep_blank_values=0, strict_parsing=0)
        #for each params in 'parse_after':
            #replace a key's value from params with a value from self.foozs, 
            #loop over this single key inserting a single value from self.fooz for each param for all fooz_objects, then continue to the next param and do the same
            #create full_param_vector var with these new values
            #construct full_path made up of: path.pathToScan - <part before '?'> + "?" + full_param_vector
            #add all 'full_path' to a dictionary named dictpath
        #print dictpath

欢迎任何帮助。谢谢

Answer 1

这样的事情可以解决问题，但是，我仍然无法解析你的问题

from collections import defaultdict
import urllib
import urlparse

# parse the url into parts
parsed = urlparse.urlparse('http://somesite.com/blog/posting/?id=6&name=Bill')

# and parse the query string into a dictionary
qs = urlparse.parse_qs(parsed.query, keep_blank_values=0, strict_parsing=0)

# this makes a new dictionary, with same keys, but all values changed to "foobar"
foozsified = { i: 'foobar' for i in qs }

# make them back to a query string: id=foobar&name=foobar
quoted = urllib.urlencode(foozsified, doseq=True)

# the original parsed result is a named tuple and cannot be changed,
# make it into a list
parsed = list(parsed)

# replace the 4th element - the query string with our new
parsed[4] = quoted

# and unparse it into a full url    
print(urlparse.urlunparse(parsed))

打印

http://somesite.com/blog/posting/?id=foobar&name=foobar

因此您可以对qs词典进行任何修改，然后使用urlunparse返回完整的网址。

使用urlparse库迭代url param

1 个答案: