How to get data from an XML file with xmlns in root
number.xml
<?xml version="1.0" encoding="utf-8"?>
<ResponseSent>
<ResponseDate xmlns="http://example.com/schema">
<emailid>123@test.com</emailid>
<number>22</number>
<sent>2017-12-05</sent>
</ResponseDate>
number.py
import xml.etree.ElementTree as ET
tree = ET.parse('number.xml')
root = tree.getroot()
for country in root.findall('ResponseDate'):
rank = country.find('emailid').text
name = country.find('number').text
print(name, rank)
Returning empty results, but when I modify the xml to name= instead of xmlns= then it is working. But, how to make this script work with the xmlns.?
python xml tree elementtree
asked Feb 8 '18 at 0:11
PrimePrime
1,96341634
add a comment |
number.xml
<?xml version="1.0" encoding="utf-8"?>
<ResponseSent>
<ResponseDate xmlns="http://example.com/schema">
<emailid>123@test.com</emailid>
<number>22</number>
<sent>2017-12-05</sent>
</ResponseDate>
number.py
import xml.etree.ElementTree as ET
tree = ET.parse('number.xml')
root = tree.getroot()
for country in root.findall('ResponseDate'):
rank = country.find('emailid').text
name = country.find('number').text
print(name, rank)
Returning empty results, but when I modify the xml to name= instead of xmlns= then it is working. But, how to make this script work with the xmlns.?
python xml tree elementtree
asked Feb 8 '18 at 0:11
PrimePrime
1,96341634
add a comment |
number.xml
<?xml version="1.0" encoding="utf-8"?>
<ResponseSent>
<ResponseDate xmlns="http://example.com/schema">
<emailid>123@test.com</emailid>
<number>22</number>
<sent>2017-12-05</sent>
</ResponseDate>
number.py
import xml.etree.ElementTree as ET
tree = ET.parse('number.xml')
root = tree.getroot()
for country in root.findall('ResponseDate'):
rank = country.find('emailid').text
name = country.find('number').text
print(name, rank)
Returning empty results, but when I modify the xml to name= instead of xmlns= then it is working. But, how to make this script work with the xmlns.?
python xml tree elementtree
asked Feb 8 '18 at 0:11
PrimePrime
1,96341634
number.xml
<?xml version="1.0" encoding="utf-8"?>
<ResponseSent>
<ResponseDate xmlns="http://example.com/schema">
<emailid>123@test.com</emailid>
<number>22</number>
<sent>2017-12-05</sent>
</ResponseDate>
number.py
import xml.etree.ElementTree as ET
tree = ET.parse('number.xml')
root = tree.getroot()
for country in root.findall('ResponseDate'):
rank = country.find('emailid').text
name = country.find('number').text
print(name, rank)
Returning empty results, but when I modify the xml to name= instead of xmlns= then it is working. But, how to make this script work with the xmlns.?
python xml tree elementtree
python xml tree elementtree
asked Feb 8 '18 at 0:11
PrimePrime
1,96341634
asked Feb 8 '18 at 0:11
PrimePrime
1,96341634
asked Feb 8 '18 at 0:11
PrimePrime
1,96341634
asked Feb 8 '18 at 0:11
PrimePrime
1,96341634
asked Feb 8 '18 at 0:11
PrimePrime
1,96341634
1,96341634
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Notice that xmlns without prefix in XML declares default namespace, and descendants element without prefix inherit default namespace from ancestor implicitly. Now to find element in namespace you can
define a prefix that references the namespace URI, and use combination of that prefix and target element's local name :
....
ns = 'd': 'http://example.com/schema'
for country in root.findall('d:ResponseDate', ns):
rank = country.find('d:emailid', ns).text
name = country.find('d:number', ns).text
print(name, rank)
answered Feb 8 '18 at 5:50
har07har07
74.9k125087
Thanks for clarification and solution.
– Prime
Feb 11 '18 at 3:23
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Notice that xmlns without prefix in XML declares default namespace, and descendants element without prefix inherit default namespace from ancestor implicitly. Now to find element in namespace you can
define a prefix that references the namespace URI, and use combination of that prefix and target element's local name :
....
ns = 'd': 'http://example.com/schema'
for country in root.findall('d:ResponseDate', ns):
rank = country.find('d:emailid', ns).text
name = country.find('d:number', ns).text
print(name, rank)
answered Feb 8 '18 at 5:50
har07har07
74.9k125087
Thanks for clarification and solution.
– Prime
Feb 11 '18 at 3:23
add a comment |
Notice that xmlns without prefix in XML declares default namespace, and descendants element without prefix inherit default namespace from ancestor implicitly. Now to find element in namespace you can
define a prefix that references the namespace URI, and use combination of that prefix and target element's local name :
....
ns = 'd': 'http://example.com/schema'
for country in root.findall('d:ResponseDate', ns):
rank = country.find('d:emailid', ns).text
name = country.find('d:number', ns).text
print(name, rank)
answered Feb 8 '18 at 5:50
har07har07
74.9k125087
Thanks for clarification and solution.
– Prime
Feb 11 '18 at 3:23
add a comment |
Notice that xmlns without prefix in XML declares default namespace, and descendants element without prefix inherit default namespace from ancestor implicitly. Now to find element in namespace you can
define a prefix that references the namespace URI, and use combination of that prefix and target element's local name :
....
ns = 'd': 'http://example.com/schema'
for country in root.findall('d:ResponseDate', ns):
rank = country.find('d:emailid', ns).text
name = country.find('d:number', ns).text
print(name, rank)
answered Feb 8 '18 at 5:50
har07har07
74.9k125087
Notice that xmlns without prefix in XML declares default namespace, and descendants element without prefix inherit default namespace from ancestor implicitly. Now to find element in namespace you can
define a prefix that references the namespace URI, and use combination of that prefix and target element's local name :
....
ns = 'd': 'http://example.com/schema'
for country in root.findall('d:ResponseDate', ns):
rank = country.find('d:emailid', ns).text
name = country.find('d:number', ns).text
print(name, rank)
answered Feb 8 '18 at 5:50
har07har07
74.9k125087
answered Feb 8 '18 at 5:50
har07har07
74.9k125087
answered Feb 8 '18 at 5:50
har07har07
74.9k125087
answered Feb 8 '18 at 5:50
har07har07
74.9k125087
74.9k125087
Thanks for clarification and solution.
– Prime
Feb 11 '18 at 3:23
add a comment |
Thanks for clarification and solution.
– Prime
Feb 11 '18 at 3:23
Thanks for clarification and solution.
– Prime
Feb 11 '18 at 3:23
Thanks for clarification and solution.
– Prime
Feb 11 '18 at 3:23
add a comment |
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