Parsing nested XML in Python?
0
I have the following goodreads response:
<GoodreadsResponse>
<Request>
</Request>
<book>
<popular_shelves>
<shelf name="test" other="1"/>
<shelf name="test2" other="2"/>
</popular_shelves/>
</book>
</GoodreadsResponse>
I want to retrieve the popular_shelves 2nd shelf item. (index 1).
Attempt 1:
from xml.etree import ElementTree as ET
root = ET.parse(urllib.urlopen(baseEndpoint+bookName)).getroot()
for atype in root.findall('book/popular_shelves'):
print(atype.get('shelf'))
Attempt 2:
genre = root.find('book').findall('popular_shelves')[0].findall('shelf')
print genre[0].text
python xml
asked Nov 25 '18 at 19:07
John LippsonJohn Lippson
312213
add a comment |
0
I have the following goodreads response:
<GoodreadsResponse>
<Request>
</Request>
<book>
<popular_shelves>
<shelf name="test" other="1"/>
<shelf name="test2" other="2"/>
</popular_shelves/>
</book>
</GoodreadsResponse>
I want to retrieve the popular_shelves 2nd shelf item. (index 1).
Attempt 1:
from xml.etree import ElementTree as ET
root = ET.parse(urllib.urlopen(baseEndpoint+bookName)).getroot()
for atype in root.findall('book/popular_shelves'):
print(atype.get('shelf'))
Attempt 2:
genre = root.find('book').findall('popular_shelves')[0].findall('shelf')
print genre[0].text
python xml
asked Nov 25 '18 at 19:07
John LippsonJohn Lippson
312213
What did you get with those attempts?
– usr2564301
Nov 25 '18 at 19:29
add a comment |
0
0
0
I have the following goodreads response:
<GoodreadsResponse>
<Request>
</Request>
<book>
<popular_shelves>
<shelf name="test" other="1"/>
<shelf name="test2" other="2"/>
</popular_shelves/>
</book>
</GoodreadsResponse>
I want to retrieve the popular_shelves 2nd shelf item. (index 1).
Attempt 1:
from xml.etree import ElementTree as ET
root = ET.parse(urllib.urlopen(baseEndpoint+bookName)).getroot()
for atype in root.findall('book/popular_shelves'):
print(atype.get('shelf'))
Attempt 2:
genre = root.find('book').findall('popular_shelves')[0].findall('shelf')
print genre[0].text
python xml
asked Nov 25 '18 at 19:07
John LippsonJohn Lippson
312213
I have the following goodreads response:
<GoodreadsResponse>
<Request>
</Request>
<book>
<popular_shelves>
<shelf name="test" other="1"/>
<shelf name="test2" other="2"/>
</popular_shelves/>
</book>
</GoodreadsResponse>
I want to retrieve the popular_shelves 2nd shelf item. (index 1).
Attempt 1:
from xml.etree import ElementTree as ET
root = ET.parse(urllib.urlopen(baseEndpoint+bookName)).getroot()
for atype in root.findall('book/popular_shelves'):
print(atype.get('shelf'))
Attempt 2:
genre = root.find('book').findall('popular_shelves')[0].findall('shelf')
print genre[0].text
python xml
python xml
asked Nov 25 '18 at 19:07
John LippsonJohn Lippson
312213
asked Nov 25 '18 at 19:07
John LippsonJohn Lippson
312213
asked Nov 25 '18 at 19:07
John LippsonJohn Lippson
312213
asked Nov 25 '18 at 19:07
John LippsonJohn Lippson
312213
asked Nov 25 '18 at 19:07
John LippsonJohn Lippson
312213
312213
What did you get with those attempts?
– usr2564301
Nov 25 '18 at 19:29
add a comment |
What did you get with those attempts?
– usr2564301
Nov 25 '18 at 19:29
What did you get with those attempts?
– usr2564301
Nov 25 '18 at 19:29
What did you get with those attempts?
– usr2564301
Nov 25 '18 at 19:29
add a comment |
2 Answers
2
active
oldest
votes
1
This is how I got the 2nd shelf item from popular_shelves:
import xml.etree.ElementTree as ET
payload = '''
<GoodreadsResponse>
<Request>
</Request>
<book>
<popular_shelves>
<shelf name="test" other="1"/>
<shelf name="test2" other="2"/>
</popular_shelves>
</book>
</GoodreadsResponse>
'''
root = ET.fromstring(payload)
shelves = root.findall("./book/popular_shelves/shelf") # this will get you the list of shelves
print shelves[1].get('name') # fetching the name of 2nd shelf item
So, we can load all shelf items under ./book/popular_shelves into a list. And, then use the list index to access 1st, 2nd, etc. shelf items.
answered Nov 25 '18 at 21:18
singletonsingleton
814
add a comment |
0
you can try module untangle, it simple and easy to work with, for example:
In [95]: from untangle import parse
In [96]: payload = '''
...: <GoodreadsResponse>
...: <Request>
...: </Request>
...: <book>
...: <popular_shelves>
...: <shelf name="test" other="1"/>
...: <shelf name="test2" other="2"/>
...: </popular_shelves>
...: </book>
...: </GoodreadsResponse>
...: '''
In [97]: obj = parse(payload)
In [98]: shelf1 = obj.GoodreadsResponse.book.popular_shelves.shelf[1]
In [99]: vars(shelf1)
Out[99]:
{'_attributes': {u'name': u'test2', u'other': u'2'},
'_name': u'shelf',
'cdata': '',
'children': ,
'is_root': False}
answered Dec 3 '18 at 9:46
panda912panda912
463
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
This is how I got the 2nd shelf item from popular_shelves:
import xml.etree.ElementTree as ET
payload = '''
<GoodreadsResponse>
<Request>
</Request>
<book>
<popular_shelves>
<shelf name="test" other="1"/>
<shelf name="test2" other="2"/>
</popular_shelves>
</book>
</GoodreadsResponse>
'''
root = ET.fromstring(payload)
shelves = root.findall("./book/popular_shelves/shelf") # this will get you the list of shelves
print shelves[1].get('name') # fetching the name of 2nd shelf item
So, we can load all shelf items under ./book/popular_shelves into a list. And, then use the list index to access 1st, 2nd, etc. shelf items.
answered Nov 25 '18 at 21:18
singletonsingleton
814
add a comment |
1
This is how I got the 2nd shelf item from popular_shelves:
import xml.etree.ElementTree as ET
payload = '''
<GoodreadsResponse>
<Request>
</Request>
<book>
<popular_shelves>
<shelf name="test" other="1"/>
<shelf name="test2" other="2"/>
</popular_shelves>
</book>
</GoodreadsResponse>
'''
root = ET.fromstring(payload)
shelves = root.findall("./book/popular_shelves/shelf") # this will get you the list of shelves
print shelves[1].get('name') # fetching the name of 2nd shelf item
So, we can load all shelf items under ./book/popular_shelves into a list. And, then use the list index to access 1st, 2nd, etc. shelf items.
answered Nov 25 '18 at 21:18
singletonsingleton
814
add a comment |
1
1
1
This is how I got the 2nd shelf item from popular_shelves:
import xml.etree.ElementTree as ET
payload = '''
<GoodreadsResponse>
<Request>
</Request>
<book>
<popular_shelves>
<shelf name="test" other="1"/>
<shelf name="test2" other="2"/>
</popular_shelves>
</book>
</GoodreadsResponse>
'''
root = ET.fromstring(payload)
shelves = root.findall("./book/popular_shelves/shelf") # this will get you the list of shelves
print shelves[1].get('name') # fetching the name of 2nd shelf item
So, we can load all shelf items under ./book/popular_shelves into a list. And, then use the list index to access 1st, 2nd, etc. shelf items.
answered Nov 25 '18 at 21:18
singletonsingleton
814
This is how I got the 2nd shelf item from popular_shelves:
import xml.etree.ElementTree as ET
payload = '''
<GoodreadsResponse>
<Request>
</Request>
<book>
<popular_shelves>
<shelf name="test" other="1"/>
<shelf name="test2" other="2"/>
</popular_shelves>
</book>
</GoodreadsResponse>
'''
root = ET.fromstring(payload)
shelves = root.findall("./book/popular_shelves/shelf") # this will get you the list of shelves
print shelves[1].get('name') # fetching the name of 2nd shelf item
So, we can load all shelf items under ./book/popular_shelves into a list. And, then use the list index to access 1st, 2nd, etc. shelf items.
answered Nov 25 '18 at 21:18
singletonsingleton
814
answered Nov 25 '18 at 21:18
singletonsingleton
814
answered Nov 25 '18 at 21:18
singletonsingleton
814
answered Nov 25 '18 at 21:18
singletonsingleton
814
814
add a comment |
add a comment |
0
you can try module untangle, it simple and easy to work with, for example:
In [95]: from untangle import parse
In [96]: payload = '''
...: <GoodreadsResponse>
...: <Request>
...: </Request>
...: <book>
...: <popular_shelves>
...: <shelf name="test" other="1"/>
...: <shelf name="test2" other="2"/>
...: </popular_shelves>
...: </book>
...: </GoodreadsResponse>
...: '''
In [97]: obj = parse(payload)
In [98]: shelf1 = obj.GoodreadsResponse.book.popular_shelves.shelf[1]
In [99]: vars(shelf1)
Out[99]:
{'_attributes': {u'name': u'test2', u'other': u'2'},
'_name': u'shelf',
'cdata': '',
'children': ,
'is_root': False}
answered Dec 3 '18 at 9:46
panda912panda912
463
add a comment |
0
you can try module untangle, it simple and easy to work with, for example:
In [95]: from untangle import parse
In [96]: payload = '''
...: <GoodreadsResponse>
...: <Request>
...: </Request>
...: <book>
...: <popular_shelves>
...: <shelf name="test" other="1"/>
...: <shelf name="test2" other="2"/>
...: </popular_shelves>
...: </book>
...: </GoodreadsResponse>
...: '''
In [97]: obj = parse(payload)
In [98]: shelf1 = obj.GoodreadsResponse.book.popular_shelves.shelf[1]
In [99]: vars(shelf1)
Out[99]:
{'_attributes': {u'name': u'test2', u'other': u'2'},
'_name': u'shelf',
'cdata': '',
'children': ,
'is_root': False}
answered Dec 3 '18 at 9:46
panda912panda912
463
add a comment |
0
0
0
you can try module untangle, it simple and easy to work with, for example:
In [95]: from untangle import parse
In [96]: payload = '''
...: <GoodreadsResponse>
...: <Request>
...: </Request>
...: <book>
...: <popular_shelves>
...: <shelf name="test" other="1"/>
...: <shelf name="test2" other="2"/>
...: </popular_shelves>
...: </book>
...: </GoodreadsResponse>
...: '''
In [97]: obj = parse(payload)
In [98]: shelf1 = obj.GoodreadsResponse.book.popular_shelves.shelf[1]
In [99]: vars(shelf1)
Out[99]:
{'_attributes': {u'name': u'test2', u'other': u'2'},
'_name': u'shelf',
'cdata': '',
'children': ,
'is_root': False}
answered Dec 3 '18 at 9:46
panda912panda912
463
you can try module untangle, it simple and easy to work with, for example:
In [95]: from untangle import parse
In [96]: payload = '''
...: <GoodreadsResponse>
...: <Request>
...: </Request>
...: <book>
...: <popular_shelves>
...: <shelf name="test" other="1"/>
...: <shelf name="test2" other="2"/>
...: </popular_shelves>
...: </book>
...: </GoodreadsResponse>
...: '''
In [97]: obj = parse(payload)
In [98]: shelf1 = obj.GoodreadsResponse.book.popular_shelves.shelf[1]
In [99]: vars(shelf1)
Out[99]:
{'_attributes': {u'name': u'test2', u'other': u'2'},
'_name': u'shelf',
'cdata': '',
'children': ,
'is_root': False}
answered Dec 3 '18 at 9:46
panda912panda912
463
answered Dec 3 '18 at 9:46
panda912panda912
463
answered Dec 3 '18 at 9:46
panda912panda912
463
answered Dec 3 '18 at 9:46
panda912panda912
463
463
add a comment |
add a comment |
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What did you get with those attempts?
– usr2564301
Nov 25 '18 at 19:29