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validating a xml doc in groovy
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
The Ask Question Wizard is Live!
Data science time! April 2019 and salary with experience
Should we burninate the [wrap] tag?Scala vs. Groovy vs. ClojureWhat does <![CDATA[]]> in XML mean?How do you parse and process HTML/XML in PHP?XML with XSD JAVAXML syntax validation in JavaXSD validation based on XML contentcvc-elt.1: Cannot find the declaration of element 'staff' - Cannot validate xml against a xsdXML Validation with XML-Reader in PHPJava / Groovy : Find XML node by Line numberHow to validate each xml tag against xsd and write the tags that pass validation to a new xml file
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I have a xml object in Groovy.. that has already been parsed by the time I get to it
def doc = new XmlSlurper().parse('sample.xml')
I want to validate it against an XSD
However in the example code for this involves the xml being presented in string or file
def xsdLocation = 'defn.xsd'
SchemaFactory.newInstance( XMLConstants.W3C_XML_SCHEMA_NS_URI)
.newSchema( new File(xsdLocation))
.newValidator()
.validate( doc )
I can't figure out what conversion or combination of StreamSource type object I need to pass in what I have (result of XmlSlurper.parse) to validate()
xml groovy xsd
asked Mar 8 at 16:16
GHZGHZ
1,67121726
add a comment |
I have a xml object in Groovy.. that has already been parsed by the time I get to it
def doc = new XmlSlurper().parse('sample.xml')
I want to validate it against an XSD
However in the example code for this involves the xml being presented in string or file
def xsdLocation = 'defn.xsd'
SchemaFactory.newInstance( XMLConstants.W3C_XML_SCHEMA_NS_URI)
.newSchema( new File(xsdLocation))
.newValidator()
.validate( doc )
I can't figure out what conversion or combination of StreamSource type object I need to pass in what I have (result of XmlSlurper.parse) to validate()
xml groovy xsd
asked Mar 8 at 16:16
GHZGHZ
1,67121726
add a comment |
I have a xml object in Groovy.. that has already been parsed by the time I get to it
def doc = new XmlSlurper().parse('sample.xml')
I want to validate it against an XSD
However in the example code for this involves the xml being presented in string or file
def xsdLocation = 'defn.xsd'
SchemaFactory.newInstance( XMLConstants.W3C_XML_SCHEMA_NS_URI)
.newSchema( new File(xsdLocation))
.newValidator()
.validate( doc )
I can't figure out what conversion or combination of StreamSource type object I need to pass in what I have (result of XmlSlurper.parse) to validate()
xml groovy xsd
asked Mar 8 at 16:16
GHZGHZ
1,67121726
I have a xml object in Groovy.. that has already been parsed by the time I get to it
def doc = new XmlSlurper().parse('sample.xml')
I want to validate it against an XSD
However in the example code for this involves the xml being presented in string or file
def xsdLocation = 'defn.xsd'
SchemaFactory.newInstance( XMLConstants.W3C_XML_SCHEMA_NS_URI)
.newSchema( new File(xsdLocation))
.newValidator()
.validate( doc )
I can't figure out what conversion or combination of StreamSource type object I need to pass in what I have (result of XmlSlurper.parse) to validate()
xml groovy xsd
xml groovy xsd
asked Mar 8 at 16:16
GHZGHZ
1,67121726
asked Mar 8 at 16:16
GHZGHZ
1,67121726
asked Mar 8 at 16:16
GHZGHZ
1,67121726
asked Mar 8 at 16:16
GHZGHZ
1,67121726
asked Mar 8 at 16:16
GHZGHZ
1,67121726
1,67121726
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
This worked for me.
import groovy.xml.XmlUtil
def doc = new XmlSlurper().parse('sample.xml')
def xsdLocation = 'defn.xsd'
SchemaFactory.newInstance( XMLConstants.W3C_XML_SCHEMA_NS_URI)
.newSchema( new File(xsdLocation))
.newValidator()
.validate(new StreamSource(new StringReader( XmlUtil.serialize(doc))))
answered Mar 8 at 16:51
GHZGHZ
1,67121726
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This worked for me.
import groovy.xml.XmlUtil
def doc = new XmlSlurper().parse('sample.xml')
def xsdLocation = 'defn.xsd'
SchemaFactory.newInstance( XMLConstants.W3C_XML_SCHEMA_NS_URI)
.newSchema( new File(xsdLocation))
.newValidator()
.validate(new StreamSource(new StringReader( XmlUtil.serialize(doc))))
answered Mar 8 at 16:51
GHZGHZ
1,67121726
add a comment |
This worked for me.
import groovy.xml.XmlUtil
def doc = new XmlSlurper().parse('sample.xml')
def xsdLocation = 'defn.xsd'
SchemaFactory.newInstance( XMLConstants.W3C_XML_SCHEMA_NS_URI)
.newSchema( new File(xsdLocation))
.newValidator()
.validate(new StreamSource(new StringReader( XmlUtil.serialize(doc))))
answered Mar 8 at 16:51
GHZGHZ
1,67121726
add a comment |
This worked for me.
import groovy.xml.XmlUtil
def doc = new XmlSlurper().parse('sample.xml')
def xsdLocation = 'defn.xsd'
SchemaFactory.newInstance( XMLConstants.W3C_XML_SCHEMA_NS_URI)
.newSchema( new File(xsdLocation))
.newValidator()
.validate(new StreamSource(new StringReader( XmlUtil.serialize(doc))))
answered Mar 8 at 16:51
GHZGHZ
1,67121726
This worked for me.
import groovy.xml.XmlUtil
def doc = new XmlSlurper().parse('sample.xml')
def xsdLocation = 'defn.xsd'
SchemaFactory.newInstance( XMLConstants.W3C_XML_SCHEMA_NS_URI)
.newSchema( new File(xsdLocation))
.newValidator()
.validate(new StreamSource(new StringReader( XmlUtil.serialize(doc))))
answered Mar 8 at 16:51
GHZGHZ
1,67121726
answered Mar 8 at 16:51
GHZGHZ
1,67121726
answered Mar 8 at 16:51
GHZGHZ
1,67121726
answered Mar 8 at 16:51
GHZGHZ
1,67121726
1,67121726
add a comment |
add a comment |
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