Convert JSON to XML in Python

Question

I see a number of questions on SO asking about ways to convert XML to JSON, but I'm interested in going the other way. Is there a python library for converting JSON to XML?

Edit: Nothing came back right away, so I went ahead and wrote a script that solves this problem.

Python already allows you to convert from JSON into a native dict (using json or, in versions < 2.6, simplejson), so I wrote a library that converts native dicts into an XML string.

https://github.com/quandyfactory/dict2xml

It supports int, float, boolean, string (and unicode), array and dict data types and arbitrary nesting (yay recursion).

I'll post this as an answer once 8 hours have passed.

There is no un step process for this conversion, but won't be hard to do using json package, and nay XML processing package. At this very site (http://stackoverflow.com/a/191617/212952) you can find some hints — Tomas Narros, Jan 24 '12 at 14:52
In java we have a nice library to convert json to xml https://github.com/javadev/underscore-java/releases — Valentyn Kolesnikov, Jan 08 '23 at 03:49

score 56 · Accepted Answer · answered Jan 24 '12 at 23:56

56

Nothing came back right away, so I went ahead and wrote a script that solves this problem.

Python already allows you to convert from JSON into a native dict (using json or, in versions < 2.6, simplejson), so I wrote a library that converts native dicts into an XML string.

https://github.com/quandyfactory/dict2xml

It supports int, float, boolean, string (and unicode), array and dict data types and arbitrary nesting (yay recursion).

answered Jan 24 '12 at 23:56

Ryan McGreal

1,323
1
9
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awesome response to actually build (and maintain!) something to resolve a SO question! – thejohnbackes Dec 11 '17 at 21:48

score 18 · Answer 2 · answered Oct 20 '13 at 06:30

If you don't have such a package, you can try:

def json2xml(json_obj, line_padding=""):
    result_list = list()

    json_obj_type = type(json_obj)

    if json_obj_type is list:
        for sub_elem in json_obj:
            result_list.append(json2xml(sub_elem, line_padding))

        return "\n".join(result_list)

    if json_obj_type is dict:
        for tag_name in json_obj:
            sub_obj = json_obj[tag_name]
            result_list.append("%s<%s>" % (line_padding, tag_name))
            result_list.append(json2xml(sub_obj, "\t" + line_padding))
            result_list.append("%s</%s>" % (line_padding, tag_name))

        return "\n".join(result_list)

    return "%s%s" % (line_padding, json_obj)

For example:

s='{"main" : {"aaa" : "10", "bbb" : [1,2,3]}}'
j = json.loads(s)
print(json2xml(j))

Result:

<main>
        <aaa>
                10
        </aaa>
        <bbb>
                1
                2
                3
        </bbb>
</main>

It is showing an Unresolved reference error on the recursive call in Python2.7 — Abhilash Cherukat, Feb 06 '15 at 11:45
given `s = {'person': {'web': {'email': 'boo@baz.com'}, 'address': {'street1': '123 Bar St', 'street2': '', 'state': 'WI', 'zip': 55555, 'city': 'Madison'}}}`. It is giving ` j = json.loads(s) File "/usr/lib/python2.7/json/__init__.py", line 338, in loads return _default_decoder.decode(s) File "/usr/lib/python2.7/json/decoder.py", line 366, in decode obj, end = self.raw_decode(s, idx=_w(s, 0).end()) File "/usr/lib/python2.7/json/decoder.py", line 382, in raw_decode obj, end = self.scan_once(s, idx) ValueError: Expecting property name: line 1 column 2 (char 1)` — Ursa Major, Jan 29 '16 at 01:13

score 15 · Answer 3 · edited May 23 '17 at 12:02

15

Load it into a dict using json.loads then use anything from this question...

Serialize Python dictionary to XML

edited May 23 '17 at 12:02

Community

1
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answered Jan 24 '12 at 17:18

wmil

3,179
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score 3 · Answer 4 · answered Sep 28 '18 at 12:04

Use dicttoxml to convert JSON directly to XML

Installation
pip install dicttoxml
or
easy_install dicttoxml

In [2]: from json import loads

In [3]: from dicttoxml import dicttoxml

In [4]: json_obj = '{"main" : {"aaa" : "10", "bbb" : [1,2,3]}}'

In [5]: xml = dicttoxml(loads(json_obj))

In [6]: print(xml)
<?xml version="1.0" encoding="UTF-8" ?><root><main type="dict"><aaa type="str">10</aaa><bbb type="list"><item type="int">1</item><item type="int">2</item><item type="int">3</item></bbb></main></root>

In [7]: xml = dicttoxml(loads(json_obj), attr_type=False)

In [8]: print(xml)
<?xml version="1.0" encoding="UTF-8" ?><root><main><aaa>10</aaa><bbb><item>1</item><item>2</item><item>3</item></bbb></main></root>

For more information on dicttoxml

this is the same library the user who asked the question wrote..and posted inside the question.......... — MichaelR, Apr 04 '21 at 14:09

score 2 · Answer 5 · answered Nov 09 '21 at 20:47

I found xmltodict to be useful. Looks like it was released after some of the posts here. https://pypi.org/project/xmltodict/

import xmltodict
import json
sample_json = {"note": {"to": "Tove", "from": "Jani", "heading": "Reminder", "body": "Don't forget me this weekend!"}}
#############
#json to xml
#############
json_to_xml = xmltodict.unparse(sample_json)
print(json_to_xml)
#############
#xmlto json
#############
x_to_j_dict = xmltodict.parse(json_to_xml)
x_to_j_string = json.dumps(x_to_j_dict)
back_to_json = json.loads(x_to_j_string)
print(back_to_json)

score 1 · Answer 6 · edited Mar 04 '19 at 12:01

1

from json import loads
from dicttoxml import dicttoxml

s='{"main" : {"aaa" : "10", "bbb" : [1,2,3]}}'
xml = dicttoxml(loads(s))

Or if your data is stored in a pandas data.frame as mine often is:

df['xml'] = df['json'].apply(lambda s: dicttoxml(json.loads(s))

edited Mar 04 '19 at 12:01

Kye

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answered Aug 03 '18 at 15:31

cangers

390
2
9

Convert JSON to XML in Python

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