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How to remove Root tag and keep rest all row tags in an xml using python
0
I've the below XML file.
<root>
<catalog>
<book id="bk101">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>44.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk102">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>45.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk103">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>46.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
</root>
I want to create another XML by eliminating the tag. So, my new XML will look like -
<catalog>
<book id="bk101">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>44.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk102">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>45.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk103">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>46.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
Below is my code and I'm able to generate byte class by eliminating the and keeping all the necessary row tags. but finally not able to convert my byte class to an xml format and getting the below error :
xml.etree.ElementTree.ParseError: junk after document element: line 11, column 0
Can you please assist?
import xml.etree.ElementTree as ET
base_tree = ET.parse('input.xml')
catalog = list(base_tree.getroot())
elemList =
for elem in catalog:
getele = ET.tostring(elem, 'utf-8')
elemList.append(getele)
byt = b''.join(elemList)
print(byt)
mytree = ET.ElementTree(ET.fromstring(byt))
dis = str(ET.tostring(mytree.getroot()), 'utf-8')
python python-3.x python-2.7
asked Nov 22 '18 at 4:18
Nabarun ChakrabortiNabarun Chakraborti
63
add a comment |
0
I've the below XML file.
<root>
<catalog>
<book id="bk101">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>44.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk102">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>45.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk103">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>46.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
</root>
I want to create another XML by eliminating the tag. So, my new XML will look like -
<catalog>
<book id="bk101">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>44.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk102">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>45.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk103">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>46.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
Below is my code and I'm able to generate byte class by eliminating the and keeping all the necessary row tags. but finally not able to convert my byte class to an xml format and getting the below error :
xml.etree.ElementTree.ParseError: junk after document element: line 11, column 0
Can you please assist?
import xml.etree.ElementTree as ET
base_tree = ET.parse('input.xml')
catalog = list(base_tree.getroot())
elemList =
for elem in catalog:
getele = ET.tostring(elem, 'utf-8')
elemList.append(getele)
byt = b''.join(elemList)
print(byt)
mytree = ET.ElementTree(ET.fromstring(byt))
dis = str(ET.tostring(mytree.getroot()), 'utf-8')
python python-3.x python-2.7
asked Nov 22 '18 at 4:18
Nabarun ChakrabortiNabarun Chakraborti
63
Your "new XML" is not well-formed XML. XML requires a root element.
– Robby Cornelissen
Nov 22 '18 at 4:20
add a comment |
0
0
0
1
I've the below XML file.
<root>
<catalog>
<book id="bk101">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>44.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk102">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>45.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk103">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>46.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
</root>
I want to create another XML by eliminating the tag. So, my new XML will look like -
<catalog>
<book id="bk101">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>44.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk102">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>45.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk103">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>46.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
Below is my code and I'm able to generate byte class by eliminating the and keeping all the necessary row tags. but finally not able to convert my byte class to an xml format and getting the below error :
xml.etree.ElementTree.ParseError: junk after document element: line 11, column 0
Can you please assist?
import xml.etree.ElementTree as ET
base_tree = ET.parse('input.xml')
catalog = list(base_tree.getroot())
elemList =
for elem in catalog:
getele = ET.tostring(elem, 'utf-8')
elemList.append(getele)
byt = b''.join(elemList)
print(byt)
mytree = ET.ElementTree(ET.fromstring(byt))
dis = str(ET.tostring(mytree.getroot()), 'utf-8')
python python-3.x python-2.7
asked Nov 22 '18 at 4:18
Nabarun ChakrabortiNabarun Chakraborti
63
I've the below XML file.
<root>
<catalog>
<book id="bk101">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>44.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk102">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>45.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk103">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>46.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
</root>
I want to create another XML by eliminating the tag. So, my new XML will look like -
<catalog>
<book id="bk101">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>44.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk102">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>45.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
<catalog>
<book id="bk103">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide</title>
<genre>Computer</genre>
<price>46.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at creating applications with XML.</description>
</book>
</catalog>
Below is my code and I'm able to generate byte class by eliminating the and keeping all the necessary row tags. but finally not able to convert my byte class to an xml format and getting the below error :
xml.etree.ElementTree.ParseError: junk after document element: line 11, column 0
Can you please assist?
import xml.etree.ElementTree as ET
base_tree = ET.parse('input.xml')
catalog = list(base_tree.getroot())
elemList =
for elem in catalog:
getele = ET.tostring(elem, 'utf-8')
elemList.append(getele)
byt = b''.join(elemList)
print(byt)
mytree = ET.ElementTree(ET.fromstring(byt))
dis = str(ET.tostring(mytree.getroot()), 'utf-8')
python python-3.x python-2.7
python python-3.x python-2.7
asked Nov 22 '18 at 4:18
Nabarun ChakrabortiNabarun Chakraborti
63
asked Nov 22 '18 at 4:18
Nabarun ChakrabortiNabarun Chakraborti
63
asked Nov 22 '18 at 4:18
Nabarun ChakrabortiNabarun Chakraborti
63
asked Nov 22 '18 at 4:18
Nabarun ChakrabortiNabarun Chakraborti
63
asked Nov 22 '18 at 4:18
Nabarun ChakrabortiNabarun Chakraborti
63
63
Your "new XML" is not well-formed XML. XML requires a root element.
– Robby Cornelissen
Nov 22 '18 at 4:20
add a comment |
Your "new XML" is not well-formed XML. XML requires a root element.
– Robby Cornelissen
Nov 22 '18 at 4:20
Your "new XML" is not well-formed XML. XML requires a root element.
– Robby Cornelissen
Nov 22 '18 at 4:20
Your "new XML" is not well-formed XML. XML requires a root element.
– Robby Cornelissen
Nov 22 '18 at 4:20
add a comment |
2 Answers
2
active
oldest
votes
1
You can use list for this.
with open('input.xml') as input_file:
text = input_file.read()
catalog = list(ET.fromstring(text))[0]
ET.tostring(catalog, encoding='utf8', method='xml')
Though resulting string will not be a valid XML.
answered Nov 22 '18 at 5:16
shoonya ekshoonya ek
365
add a comment |
0
root element is mandatory for being XML.
For just text processing maybe we could just do
import re
pattern = re.compile("<[/]{0,1}root>")
removed = re.sub(pattern, '', "<root>something</root>");
print(removed)
?
answered Nov 22 '18 at 4:36
suplsupl
745
But then how will you solve the second problem ? regex.info/blog/2006-09-15/247 :). I would avoid regex when I have some structure like XML.
– 0xc0de
Nov 22 '18 at 4:51
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
You can use list for this.
with open('input.xml') as input_file:
text = input_file.read()
catalog = list(ET.fromstring(text))[0]
ET.tostring(catalog, encoding='utf8', method='xml')
Though resulting string will not be a valid XML.
answered Nov 22 '18 at 5:16
shoonya ekshoonya ek
365
add a comment |
1
You can use list for this.
with open('input.xml') as input_file:
text = input_file.read()
catalog = list(ET.fromstring(text))[0]
ET.tostring(catalog, encoding='utf8', method='xml')
Though resulting string will not be a valid XML.
answered Nov 22 '18 at 5:16
shoonya ekshoonya ek
365
add a comment |
1
1
1
You can use list for this.
with open('input.xml') as input_file:
text = input_file.read()
catalog = list(ET.fromstring(text))[0]
ET.tostring(catalog, encoding='utf8', method='xml')
Though resulting string will not be a valid XML.
answered Nov 22 '18 at 5:16
shoonya ekshoonya ek
365
You can use list for this.
with open('input.xml') as input_file:
text = input_file.read()
catalog = list(ET.fromstring(text))[0]
ET.tostring(catalog, encoding='utf8', method='xml')
Though resulting string will not be a valid XML.
answered Nov 22 '18 at 5:16
shoonya ekshoonya ek
365
answered Nov 22 '18 at 5:16
shoonya ekshoonya ek
365
answered Nov 22 '18 at 5:16
shoonya ekshoonya ek
365
answered Nov 22 '18 at 5:16
shoonya ekshoonya ek
365
365
add a comment |
add a comment |
0
root element is mandatory for being XML.
For just text processing maybe we could just do
import re
pattern = re.compile("<[/]{0,1}root>")
removed = re.sub(pattern, '', "<root>something</root>");
print(removed)
?
answered Nov 22 '18 at 4:36
suplsupl
745
But then how will you solve the second problem ? regex.info/blog/2006-09-15/247 :). I would avoid regex when I have some structure like XML.
– 0xc0de
Nov 22 '18 at 4:51
add a comment |
0
root element is mandatory for being XML.
For just text processing maybe we could just do
import re
pattern = re.compile("<[/]{0,1}root>")
removed = re.sub(pattern, '', "<root>something</root>");
print(removed)
?
answered Nov 22 '18 at 4:36
suplsupl
745
But then how will you solve the second problem ? regex.info/blog/2006-09-15/247 :). I would avoid regex when I have some structure like XML.
– 0xc0de
Nov 22 '18 at 4:51
add a comment |
0
0
0
root element is mandatory for being XML.
For just text processing maybe we could just do
import re
pattern = re.compile("<[/]{0,1}root>")
removed = re.sub(pattern, '', "<root>something</root>");
print(removed)
?
answered Nov 22 '18 at 4:36
suplsupl
745
root element is mandatory for being XML.
For just text processing maybe we could just do
import re
pattern = re.compile("<[/]{0,1}root>")
removed = re.sub(pattern, '', "<root>something</root>");
print(removed)
?
answered Nov 22 '18 at 4:36
suplsupl
745
answered Nov 22 '18 at 4:36
suplsupl
745
answered Nov 22 '18 at 4:36
suplsupl
745
answered Nov 22 '18 at 4:36
suplsupl
745
745
But then how will you solve the second problem ? regex.info/blog/2006-09-15/247 :). I would avoid regex when I have some structure like XML.
– 0xc0de
Nov 22 '18 at 4:51
add a comment |
But then how will you solve the second problem ? regex.info/blog/2006-09-15/247 :). I would avoid regex when I have some structure like XML.
– 0xc0de
Nov 22 '18 at 4:51
But then how will you solve the second problem ? regex.info/blog/2006-09-15/247 :). I would avoid regex when I have some structure like XML.
– 0xc0de
Nov 22 '18 at 4:51
But then how will you solve the second problem ? regex.info/blog/2006-09-15/247 :). I would avoid regex when I have some structure like XML.
– 0xc0de
Nov 22 '18 at 4:51
add a comment |
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Your "new XML" is not well-formed XML. XML requires a root element.
– Robby Cornelissen
Nov 22 '18 at 4:20