I am using python 3.6.3a. I would like to generate payload for each of the json records. I am using each variable to access the record. How to assign variable value (each in this case) in payload? I tried {each} and other methods but didn't work.
code snippet below.
json_records = [{"description":"<p>This is scenario1<\/p>","owner":"deb",
"priority":"high"},
{"description":"<p>This is scenario2<\/p>","owner":"deb",
"priority":"medium"}]
json_object = json.loads(json_records)
for each in json_object:
payload = """
{
"subject": "test",
"fieldValues": [
{each}
]
}
"""
There are two ways to approach this problem.
One way could be creating a dict() object and inserting keys as you wish, then json.dumps(object) to convert into string payload as in:
import json
json_records = [{"description":"This is scenario1</p>","owner":"deb","priority":"high"}
,{"description":"This is scenario2</p>","owner":"deb","priority":"medium"}]
for obj in json_records:
payload = dict()
payload['subject'] = 'test'
for key,value in obj.items():
payload['fieldName'] = {
key:value
}
print(json.dumps(payload))
#{"subject": "test", "fieldName": {"priority": "high"}}
#{"subject": "test", "fieldName": {"priority": "medium"}}
Second way is to create a textual payload from string as in, however if you need a valid JSON at the end, this would require a post-step of validation (something like try json.loads(payload) - So I'd just use the first method. I would use this method only if I have a specific requirements to generate the payload in a certain way.
import json
json_records = [{"description":"This is scenario1</p>","owner":"deb","priority":"high"}
,{"description":"This is scenario2</p>","owner":"deb","priority":"medium"}]
# json_object = json.loads(json_records) # json.loads works only on byte-like strings. your object is already in python in this case.
for obj in json_records:
payload = """
{
"subject": "test",
"fieldValues": [
%s
]
}
""" % (obj["priority"])
print(payload)
#{
# "subject": "test",
# "fieldValues": [
# high
# ]
# }
#
#
# {
# "subject": "test",
# "fieldValues": [
# medium
# ]
# }
You could make payload a Template string and use it to put the data in each JSON record into the format you want. Bracket {} characters have not special meaning in Templates, which is what makes using them easy.
Doing that will create a valid string representation of a dictionary containing everything. You can turn this into an actual Python dictionary data-structure using the ast.literal_eval() function, and then convert that into JSON string format — which I think is the final format you're after.
rom ast import literal_eval
import json
from string import Template
from textwrap import dedent
json_records = '''[{"description":"<p>This is scenario1<\/p>","owner":"deb",
"priority":"high"},
{"description":"<p>This is scenario2<\/p>","owner":"deb",
"priority":"medium"}]'''
json_object = json.loads(json_records)
payload = Template(dedent("""
{
"subject": "test",
"fieldValues": [
$each
]
}""")
)
for each in json_object:
obj = literal_eval(payload.substitute(dict(each=each)))
print(json.dumps(obj, indent=2))
Output:
{
"subject": "test",
"fieldValues": [
{
"description": "<p>This is scenario1</p>",
"owner": "deb",
"priority": "high"
}
]
}
{
"subject": "test",
"fieldValues": [
{
"description": "<p>This is scenario2</p>",
"owner": "deb",
"priority": "medium"
}
]
}
I have a JSON file where I need to replace the UUID and update it with another one. I'm having trouble replacing the deeply nested keys and values.
Below is my JSON file that I need to read in python, replace the keys and values and update the file.
JSON file - myfile.json
{
"name": "Shipping box"
"company":"Detla shipping"
"description":"---"
"details" : {
"boxes":[
{
"box_name":"alpha",
"id":"a3954710-5075-4f52-8eb4-1137be51bf14"
},
{
"box_name":"beta",
"id":"31be3763-3d63-4e70-a9b6-d197b5cb6929"
}
]
}
"container": [
"a3954710-5075-4f52-8eb4-1137be51bf14":[],
"31be3763-3d63-4e70-a9b6-d197b5cb6929":[]
]
"data":[
{
"data_series":[],
"other":50
},
{
"data_series":[],
"other":40
},
{
"data_series":
{
"a3954710-5075-4f52-8eb4-1137be51bf14":
{
{
"dimentions":[2,10,12]
}
},
"31be3763-3d63-4e70-a9b6-d197b5cb6929":
{
{
"dimentions":[3,9,12]
}
}
},
"other":50
}
]
}
I want achieve something like the following-
"details" : {
"boxes":[
{
"box_name":"alpha"
"id":"replace_uuid"
},
}
.
.
.
"data":[ {
"data_series":
{
"replace_uuid":
{
{
"dimentions":[2,10,12]
}
}
]
In such a type of deeply nested dictionary, how can we replace all the occurrence of keys and values with another string, here replace_uuid?
I tried with pop() and dotty_dict but I wasn't able to replace the nested list.
I was able to achieve it in the following way-
def uuid_change(): #generate a random uuid
new_uuid = uuid.uuid4()
return str(new_uuid)
dict = json.load(f)
for uid in dict[details][boxes]:
old_id = uid['id']
replace_id = uuid_change()
uid['id'] = replace_id
for i in range(n):
for uid1 in dict['container'][i].keys()
if uid1 == old_id:
dict['container'][i][replace_id]
= dict['container'][i].pop(uid1) #replace the key
for uid2 in dict['data'][2]['data_series'].keys()
if uid2 == old_id:
dict['data'][2]['data_series'][replace_id]
= dict['data'][2]['data_series'].pop(uid2) #replace the key
following Update json nodes in Python using jsonpath, would like to know how one might update the JSON data given a certain context.
So, say we pick the exact same JSON example:
{
"SchemeId": 10,
"nominations": [
{
"nominationId": 1
}
]
}
But this time, would like to double the value of the original value, hence some lambda function is needed which takes into account the current node value.
No need for lambdas; for example, to double SchemeId, something like this should work:
data = json.loads("""the json string above""")
jsonpath_expr = parse('$.SchemeId')
jsonpath_expr.find(data)
val = jsonpath_expr.find(data)[0].value
jsonpath_expr.update(data, val*2)
print(json.dumps(data, indent=2))
Output:
{
"SchemeId": 20,
"nominations": [
{
"nominationId": 1
}
]
}
Here is example with lambda expression:
import json
from jsonpath_ng import parse
settings = '''{
"choices": {
"atm": {
"cs": "Strom",
"en": "Tree"
},
"bar": {
"cs": "Dům",
"en": "House"
},
"sea": {
"cs": "Moře",
"en": "Sea"
}
}
}'''
json_data = json.loads(settings)
pattern = parse('$.choices.*')
def magic(f: dict, to_lang='cs'):
return f[to_lang]
pattern.update(json_data,
lambda data_field, data, field: data.update({field: magic(data[field])}))
json_data
returns
{
'choices': {
'atm': 'Strom',
'bar': 'Dům',
'sea': 'Moře'
}
}
I'm in over my head, trying to parse JSON for my first time and dealing with a multi dimensional array.
{
"secret": "[Hidden]",
"minutes": 20,
"link": "http:\/\/www.1.com",
"bookmark_collection": {
"free_link": {
"name": "#free_link#",
"bookmarks": [
{
"name": "1",
"link": "http:\/\/www.1.com"
},
{
"name": "2",
"link": "http:\/\/2.dk"
},
{
"name": "3",
"link": "http:\/\/www.3.in"
}
]
},
"boarding_pass": {
"name": "Boarding Pass",
"bookmarks": [
{
"name": "1",
"link": "http:\/\/www.1.com\/"
},
{
"name": "2",
"link": "http:\/\/www.2.com\/"
},
{
"name": "3",
"link": "http:\/\/www.3.hk"
}
]
},
"sublinks": {
"name": "sublinks",
"link": [
"http:\/\/www.1.com",
"http:\/\/www.2.com",
"http:\/\/www.3.com"
]
}
}
}
This is divided into 3 parts, the static data on my first dimension (secret, minutes, link) Which i need to get as seperate strings.
Then I need a dictionary per "bookmark collection" which does not have fixed names, so I need the name of them and the links/names of each bookmark.
Then there is the seperate sublinks which is always the same, where I need all the links in a seperate dictionary.
I'm reading about parsing JSON but most of the stuff I find is a simple array put into 1 dictionary.
Does anyone have any good techniques to do this ?
After you parse the JSON, you will end up with a Python dict. So, suppose the above JSON is in a string named input_data:
import json
# This converts from JSON to a python dict
parsed_input = json.loads(input_data)
# Now, all of your static variables are referenceable as keys:
secret = parsed_input['secret']
minutes = parsed_input['minutes']
link = parsed_input['link']
# Plus, you can get your bookmark collection as:
bookmark_collection = parsed_input['bookmark_collection']
# Print a list of names of the bookmark collections...
print bookmark_collection.keys() # Note this contains sublinks, so remove it if needed
# Get the name of the Boarding Pass bookmark:
print bookmark_collection['boarding_pass']['name']
# Print out a list of all bookmark links as:
# Boarding Pass
# * 1: http://www.1.com/
# * 2: http://www.2.com/
# ...
for bookmark_definition in bookmark_collection.values():
# Skip sublinks...
if bookmark_definition['name'] == 'sublinks':
continue
print bookmark_definition['name']
for bookmark in bookmark_definition['bookmarks']:
print " * %(name)s: %(link)s" % bookmark
# Get the sublink definition:
sublinks = parsed_input['bookmark_collection']['sublinks']
# .. and print them
print sublinks['name']
for link in sublinks['link']:
print ' *', link
Hmm, doesn't json.loads do the trick?
For example, if your data is in a file,
import json
text = open('/tmp/mydata.json').read()
d = json.loads(text)
# first level fields
print d['minutes'] # or 'secret' or 'link'
# the names of each of bookmark_collections's items
print d['bookmark_collection'].keys()
# the sublinks section, as a dict
print d['bookmark_collection']['sublinks']
The output of this code (given your sample input above) is:
20
[u'sublinks', u'free_link', u'boarding_pass']
{u'link': [u'http://www.1.com', u'http://www.2.com', u'http://www.3.com'], u'name': u'sublinks'}
Which, I think, gets you what you need?
I've seen a fair share of ungainly XML->JSON code on the web, and having interacted with Stack's users for a bit, I'm convinced that this crowd can help more than the first few pages of Google results can.
So, we're parsing a weather feed, and we need to populate weather widgets on a multitude of web sites. We're looking now into Python-based solutions.
This public weather.com RSS feed is a good example of what we'd be parsing (our actual weather.com feed contains additional information because of a partnership w/them).
In a nutshell, how should we convert XML to JSON using Python?
xmltodict (full disclosure: I wrote it) can help you convert your XML to a dict+list+string structure, following this "standard". It is Expat-based, so it's very fast and doesn't need to load the whole XML tree in memory.
Once you have that data structure, you can serialize it to JSON:
import xmltodict, json
o = xmltodict.parse('<e> <a>text</a> <a>text</a> </e>')
json.dumps(o) # '{"e": {"a": ["text", "text"]}}'
There is no "one-to-one" mapping between XML and JSON, so converting one to the other necessarily requires some understanding of what you want to do with the results.
That being said, Python's standard library has several modules for parsing XML (including DOM, SAX, and ElementTree). As of Python 2.6, support for converting Python data structures to and from JSON is included in the json module.
So the infrastructure is there.
You can use the xmljson library to convert using different XML JSON conventions.
For example, this XML:
<p id="1">text</p>
translates via the BadgerFish convention into this:
{
'p': {
'#id': 1,
'$': 'text'
}
}
and via the GData convention into this (attributes are not supported):
{
'p': {
'$t': 'text'
}
}
... and via the Parker convention into this (attributes are not supported):
{
'p': 'text'
}
It's possible to convert from XML to JSON and from JSON to XML using the same
conventions:
>>> import json, xmljson
>>> from lxml.etree import fromstring, tostring
>>> xml = fromstring('<p id="1">text</p>')
>>> json.dumps(xmljson.badgerfish.data(xml))
'{"p": {"#id": 1, "$": "text"}}'
>>> xmljson.parker.etree({'ul': {'li': [1, 2]}})
# Creates [<ul><li>1</li><li>2</li></ul>]
Disclosure: I wrote this library. Hope it helps future searchers.
To anyone that may still need this. Here's a newer, simple code to do this conversion.
from xml.etree import ElementTree as ET
xml = ET.parse('FILE_NAME.xml')
parsed = parseXmlToJson(xml)
def parseXmlToJson(xml):
response = {}
for child in list(xml):
if len(list(child)) > 0:
response[child.tag] = parseXmlToJson(child)
else:
response[child.tag] = child.text or ''
# one-liner equivalent
# response[child.tag] = parseXmlToJson(child) if len(list(child)) > 0 else child.text or ''
return response
If some time you get only response code instead of all data then error like json parse will be there so u need to convert it as text
import xmltodict
data = requests.get(url)
xpars = xmltodict.parse(data.text)
json = json.dumps(xpars)
print json
Here's the code I built for that. There's no parsing of the contents, just plain conversion.
from xml.dom import minidom
import simplejson as json
def parse_element(element):
dict_data = dict()
if element.nodeType == element.TEXT_NODE:
dict_data['data'] = element.data
if element.nodeType not in [element.TEXT_NODE, element.DOCUMENT_NODE,
element.DOCUMENT_TYPE_NODE]:
for item in element.attributes.items():
dict_data[item[0]] = item[1]
if element.nodeType not in [element.TEXT_NODE, element.DOCUMENT_TYPE_NODE]:
for child in element.childNodes:
child_name, child_dict = parse_element(child)
if child_name in dict_data:
try:
dict_data[child_name].append(child_dict)
except AttributeError:
dict_data[child_name] = [dict_data[child_name], child_dict]
else:
dict_data[child_name] = child_dict
return element.nodeName, dict_data
if __name__ == '__main__':
dom = minidom.parse('data.xml')
f = open('data.json', 'w')
f.write(json.dumps(parse_element(dom), sort_keys=True, indent=4))
f.close()
There is a method to transport XML-based markup as JSON which allows it to be losslessly converted back to its original form. See http://jsonml.org/.
It's a kind of XSLT of JSON. I hope you find it helpful
I'd suggest not going for a direct conversion. Convert XML to an object, then from the object to JSON.
In my opinion, this gives a cleaner definition of how the XML and JSON correspond.
It takes time to get right and you may even write tools to help you with generating some of it, but it would look roughly like this:
class Channel:
def __init__(self)
self.items = []
self.title = ""
def from_xml( self, xml_node ):
self.title = xml_node.xpath("title/text()")[0]
for x in xml_node.xpath("item"):
item = Item()
item.from_xml( x )
self.items.append( item )
def to_json( self ):
retval = {}
retval['title'] = title
retval['items'] = []
for x in items:
retval.append( x.to_json() )
return retval
class Item:
def __init__(self):
...
def from_xml( self, xml_node ):
...
def to_json( self ):
...
You may want to have a look at http://designtheory.org/library/extrep/designdb-1.0.pdf. This project starts off with an XML to JSON conversion of a large library of XML files. There was much research done in the conversion, and the most simple intuitive XML -> JSON mapping was produced (it is described early in the document). In summary, convert everything to a JSON object, and put repeating blocks as a list of objects.
objects meaning key/value pairs (dictionary in Python, hashmap in Java, object in JavaScript)
There is no mapping back to XML to get an identical document, the reason is, it is unknown whether a key/value pair was an attribute or an <key>value</key>, therefore that information is lost.
If you ask me, attributes are a hack to start; then again they worked well for HTML.
Well, probably the simplest way is just parse the XML into dictionaries and then serialize that with simplejson.
When I do anything with XML in python I almost always use the lxml package. I suspect that most people use lxml. You could use xmltodict but you will have to pay the penalty of parsing the XML again.
To convert XML to json with lxml you:
Parse XML document with lxml
Convert lxml to a dict
Convert list to json
I use the following class in my projects. Use the toJson method.
from lxml import etree
import json
class Element:
'''
Wrapper on the etree.Element class. Extends functionality to output element
as a dictionary.
'''
def __init__(self, element):
'''
:param: element a normal etree.Element instance
'''
self.element = element
def toDict(self):
'''
Returns the element as a dictionary. This includes all child elements.
'''
rval = {
self.element.tag: {
'attributes': dict(self.element.items()),
},
}
for child in self.element:
rval[self.element.tag].update(Element(child).toDict())
return rval
class XmlDocument:
'''
Wraps lxml to provide:
- cleaner access to some common lxml.etree functions
- converter from XML to dict
- converter from XML to json
'''
def __init__(self, xml = '<empty/>', filename=None):
'''
There are two ways to initialize the XmlDocument contents:
- String
- File
You don't have to initialize the XmlDocument during instantiation
though. You can do it later with the 'set' method. If you choose to
initialize later XmlDocument will be initialized with "<empty/>".
:param: xml Set this argument if you want to parse from a string.
:param: filename Set this argument if you want to parse from a file.
'''
self.set(xml, filename)
def set(self, xml=None, filename=None):
'''
Use this to set or reset the contents of the XmlDocument.
:param: xml Set this argument if you want to parse from a string.
:param: filename Set this argument if you want to parse from a file.
'''
if filename is not None:
self.tree = etree.parse(filename)
self.root = self.tree.getroot()
else:
self.root = etree.fromstring(xml)
self.tree = etree.ElementTree(self.root)
def dump(self):
etree.dump(self.root)
def getXml(self):
'''
return document as a string
'''
return etree.tostring(self.root)
def xpath(self, xpath):
'''
Return elements that match the given xpath.
:param: xpath
'''
return self.tree.xpath(xpath);
def nodes(self):
'''
Return all elements
'''
return self.root.iter('*')
def toDict(self):
'''
Convert to a python dictionary
'''
return Element(self.root).toDict()
def toJson(self, indent=None):
'''
Convert to JSON
'''
return json.dumps(self.toDict(), indent=indent)
if __name__ == "__main__":
xml='''<system>
<product>
<demod>
<frequency value='2.215' units='MHz'>
<blah value='1'/>
</frequency>
</demod>
</product>
</system>
'''
doc = XmlDocument(xml)
print doc.toJson(indent=4)
The output from the built in main is:
{
"system": {
"attributes": {},
"product": {
"attributes": {},
"demod": {
"attributes": {},
"frequency": {
"attributes": {
"units": "MHz",
"value": "2.215"
},
"blah": {
"attributes": {
"value": "1"
}
}
}
}
}
}
}
Which is a transformation of this xml:
<system>
<product>
<demod>
<frequency value='2.215' units='MHz'>
<blah value='1'/>
</frequency>
</demod>
</product>
</system>
I found for simple XML snips, use regular expression would save troubles. For example:
# <user><name>Happy Man</name>...</user>
import re
names = re.findall(r'<name>(\w+)<\/name>', xml_string)
# do some thing to names
To do it by XML parsing, as #Dan said, there is not one-for-all solution because the data is different. My suggestion is to use lxml. Although not finished to json, lxml.objectify give quiet good results:
>>> from lxml import objectify
>>> root = objectify.fromstring("""
... <root xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
... <a attr1="foo" attr2="bar">1</a>
... <a>1.2</a>
... <b>1</b>
... <b>true</b>
... <c>what?</c>
... <d xsi:nil="true"/>
... </root>
... """)
>>> print(str(root))
root = None [ObjectifiedElement]
a = 1 [IntElement]
* attr1 = 'foo'
* attr2 = 'bar'
a = 1.2 [FloatElement]
b = 1 [IntElement]
b = True [BoolElement]
c = 'what?' [StringElement]
d = None [NoneElement]
* xsi:nil = 'true'
While the built-in libs for XML parsing are quite good I am partial to lxml.
But for parsing RSS feeds, I'd recommend Universal Feed Parser, which can also parse Atom.
Its main advantage is that it can digest even most malformed feeds.
Python 2.6 already includes a JSON parser, but a newer version with improved speed is available as simplejson.
With these tools building your app shouldn't be that difficult.
My answer addresses the specific (and somewhat common) case where you don't really need to convert the entire xml to json, but what you need is to traverse/access specific parts of the xml, and you need it to be fast, and simple (using json/dict-like operations).
Approach
For this, it is important to note that parsing an xml to etree using lxml is super fast. The slow part in most of the other answers is the second pass: traversing the etree structure (usually in python-land), converting it to json.
Which leads me to the approach I found best for this case: parsing the xml using lxml, and then wrapping the etree nodes (lazily), providing them with a dict-like interface.
Code
Here's the code:
from collections import Mapping
import lxml.etree
class ETreeDictWrapper(Mapping):
def __init__(self, elem, attr_prefix = '#', list_tags = ()):
self.elem = elem
self.attr_prefix = attr_prefix
self.list_tags = list_tags
def _wrap(self, e):
if isinstance(e, basestring):
return e
if len(e) == 0 and len(e.attrib) == 0:
return e.text
return type(self)(
e,
attr_prefix = self.attr_prefix,
list_tags = self.list_tags,
)
def __getitem__(self, key):
if key.startswith(self.attr_prefix):
return self.elem.attrib[key[len(self.attr_prefix):]]
else:
subelems = [ e for e in self.elem.iterchildren() if e.tag == key ]
if len(subelems) > 1 or key in self.list_tags:
return [ self._wrap(x) for x in subelems ]
elif len(subelems) == 1:
return self._wrap(subelems[0])
else:
raise KeyError(key)
def __iter__(self):
return iter(set( k.tag for k in self.elem) |
set( self.attr_prefix + k for k in self.elem.attrib ))
def __len__(self):
return len(self.elem) + len(self.elem.attrib)
# defining __contains__ is not necessary, but improves speed
def __contains__(self, key):
if key.startswith(self.attr_prefix):
return key[len(self.attr_prefix):] in self.elem.attrib
else:
return any( e.tag == key for e in self.elem.iterchildren() )
def xml_to_dictlike(xmlstr, attr_prefix = '#', list_tags = ()):
t = lxml.etree.fromstring(xmlstr)
return ETreeDictWrapper(
t,
attr_prefix = '#',
list_tags = set(list_tags),
)
This implementation is not complete, e.g., it doesn't cleanly support cases where an element has both text and attributes, or both text and children (only because I didn't need it when I wrote it...) It should be easy to improve it, though.
Speed
In my specific use case, where I needed to only process specific elements of the xml, this approach gave a suprising and striking speedup by a factor of 70 (!) compared to using #Martin Blech's xmltodict and then traversing the dict directly.
Bonus
As a bonus, since our structure is already dict-like, we get another alternative implementation of xml2json for free. We just need to pass our dict-like structure to json.dumps. Something like:
def xml_to_json(xmlstr, **kwargs):
x = xml_to_dictlike(xmlstr, **kwargs)
return json.dumps(x)
If your xml includes attributes, you'd need to use some alphanumeric attr_prefix (e.g. "ATTR_"), to ensure the keys are valid json keys.
I haven't benchmarked this part.
check out lxml2json (disclosure: I wrote it)
https://github.com/rparelius/lxml2json
it's very fast, lightweight (only requires lxml), and one advantage is that you have control over whether certain elements are converted to lists or dicts
jsonpickle or if you're using feedparser, you can try feed_parser_to_json.py
You can use declxml. It has advanced features like multi attributes and complex nested support. You just need to write a simple processor for it. Also with the same code, you can convert back to JSON as well. It is fairly straightforward and the documentation is awesome.
Link: https://declxml.readthedocs.io/en/latest/index.html
If you don't want to use any external libraries and 3rd party tools, Try below code.
Code
import re
import json
def getdict(content):
res=re.findall("<(?P<var>\S*)(?P<attr>[^/>]*)(?:(?:>(?P<val>.*?)</(?P=var)>)|(?:/>))",content)
if len(res)>=1:
attreg="(?P<avr>\S+?)(?:(?:=(?P<quote>['\"])(?P<avl>.*?)(?P=quote))|(?:=(?P<avl1>.*?)(?:\s|$))|(?P<avl2>[\s]+)|$)"
if len(res)>1:
return [{i[0]:[{"#attributes":[{j[0]:(j[2] or j[3] or j[4])} for j in re.findall(attreg,i[1].strip())]},{"$values":getdict(i[2])}]} for i in res]
else:
return {res[0]:[{"#attributes":[{j[0]:(j[2] or j[3] or j[4])} for j in re.findall(attreg,res[1].strip())]},{"$values":getdict(res[2])}]}
else:
return content
with open("test.xml","r") as f:
print(json.dumps(getdict(f.read().replace('\n',''))))
Sample Input
<details class="4b" count=1 boy>
<name type="firstname">John</name>
<age>13</age>
<hobby>Coin collection</hobby>
<hobby>Stamp collection</hobby>
<address>
<country>USA</country>
<state>CA</state>
</address>
</details>
<details empty="True"/>
<details/>
<details class="4a" count=2 girl>
<name type="firstname">Samantha</name>
<age>13</age>
<hobby>Fishing</hobby>
<hobby>Chess</hobby>
<address current="no">
<country>Australia</country>
<state>NSW</state>
</address>
</details>
Output
[
{
"details": [
{
"#attributes": [
{
"class": "4b"
},
{
"count": "1"
},
{
"boy": ""
}
]
},
{
"$values": [
{
"name": [
{
"#attributes": [
{
"type": "firstname"
}
]
},
{
"$values": "John"
}
]
},
{
"age": [
{
"#attributes": []
},
{
"$values": "13"
}
]
},
{
"hobby": [
{
"#attributes": []
},
{
"$values": "Coin collection"
}
]
},
{
"hobby": [
{
"#attributes": []
},
{
"$values": "Stamp collection"
}
]
},
{
"address": [
{
"#attributes": []
},
{
"$values": [
{
"country": [
{
"#attributes": []
},
{
"$values": "USA"
}
]
},
{
"state": [
{
"#attributes": []
},
{
"$values": "CA"
}
]
}
]
}
]
}
]
}
]
},
{
"details": [
{
"#attributes": [
{
"empty": "True"
}
]
},
{
"$values": ""
}
]
},
{
"details": [
{
"#attributes": []
},
{
"$values": ""
}
]
},
{
"details": [
{
"#attributes": [
{
"class": "4a"
},
{
"count": "2"
},
{
"girl": ""
}
]
},
{
"$values": [
{
"name": [
{
"#attributes": [
{
"type": "firstname"
}
]
},
{
"$values": "Samantha"
}
]
},
{
"age": [
{
"#attributes": []
},
{
"$values": "13"
}
]
},
{
"hobby": [
{
"#attributes": []
},
{
"$values": "Fishing"
}
]
},
{
"hobby": [
{
"#attributes": []
},
{
"$values": "Chess"
}
]
},
{
"address": [
{
"#attributes": [
{
"current": "no"
}
]
},
{
"$values": [
{
"country": [
{
"#attributes": []
},
{
"$values": "Australia"
}
]
},
{
"state": [
{
"#attributes": []
},
{
"$values": "NSW"
}
]
}
]
}
]
}
]
}
]
}
]
This stuff here is actively maintained and so far is my favorite: xml2json in python
I published one on github a while back..
https://github.com/davlee1972/xml_to_json
This converter is written in Python and will convert one or more XML files into JSON / JSONL files
It requires a XSD schema file to figure out nested json structures (dictionaries vs lists) and json equivalent data types.
python xml_to_json.py -x PurchaseOrder.xsd PurchaseOrder.xml
INFO - 2018-03-20 11:10:24 - Parsing XML Files..
INFO - 2018-03-20 11:10:24 - Processing 1 files
INFO - 2018-03-20 11:10:24 - Parsing files in the following order:
INFO - 2018-03-20 11:10:24 - ['PurchaseOrder.xml']
DEBUG - 2018-03-20 11:10:24 - Generating schema from PurchaseOrder.xsd
DEBUG - 2018-03-20 11:10:24 - Parsing PurchaseOrder.xml
DEBUG - 2018-03-20 11:10:24 - Writing to file PurchaseOrder.json
DEBUG - 2018-03-20 11:10:24 - Completed PurchaseOrder.xml
I also have a follow up xml to parquet converter that works in a similar fashion
https://github.com/blackrock/xml_to_parquet