How to find element by attribute value in GPath?

Question

What is an alternative to this XPath //div[@id='foo'] in GPath? In general, where I can find this documentation?

score 30 · Accepted Answer · edited Dec 10 '15 at 13:38

30

Here is the corresponding snippet:

def node = new XmlSlurper().parseText(...)
def foo = node.depthFirst().findAll { it.name() == 'div' && it.@id == 'foo'}

A few other links you may want to read:

edited Dec 10 '15 at 13:38

BugOrFeature

answered Aug 11 '11 at 04:55

Nicolas Modrzyk

4

It's Groovy, not GPath. I need a single GPath expression. – yegor256 Aug 11 '11 at 04:59
3

Not too sure what you mean, since GPath is mostly using groovy syntax with the following extra symbols: ".." that returns the parent "*" that returns all the children "**" that act as a depth first loop "@" that is used to access a property the normal node accessor. – Nicolas Modrzyk Aug 11 '11 at 05:13

score 9 · Answer 2 · answered Aug 11 '11 at 08:53

9

The previous poster gave you all that's required: Assuming your document has been slurped into xml, you want

def foo = xml.path.to.div.find{it.@id == 'foo'}

to find a single result. Or findAll to find all results.

answered Aug 11 '11 at 08:53

winstaan74

`//` in XPath means that I don't know the exact path to `div` – yegor256 Aug 12 '11 at 04:09
1

ok. then the previous poster gave you the solution: - use `xml.depthFirst().find{it.name() == 'div' && id.@id == 'foo}` – winstaan74 Aug 12 '11 at 10:00
@winstaan74 What's the expression for getting all elements with a specific attribute name? – raffian Jun 01 '14 at 02:03
2

`xml.'**'.findAll { it.@attributeName }` – AbuNassar Jul 06 '16 at 13:57

score 3 · Answer 3 · answered Apr 24 '15 at 14:58

To mimic the expression //div[@id='foo'] the closest thing you can do with a GPath is:

def xml = new XmlParser().parseText(text)
xml.'**'.div.findAll { it.@id=="foo" }

the '**' is pretty much the same as '//' in your XPath.

xml.'**'.div

will yield all the nodes of type div at any level.

Later filtering with findAll() with the given closure you get a list of nodes as you do in the XPath case

score 1 · Answer 4 · edited Jan 08 '12 at 22:23

1

what you need is this:

def root = new XmlSlurper().parseText(<locOfXmlFileYouAreParsing>.toURL().text)

def foundNode = root.'**'.find{ it.@id == "foo" }

its the double * that will let you find it without knowing the path. At least this is how I do it.

edited Jan 08 '12 at 22:23

Dan

answered Jan 08 '12 at 22:20

Grant

1

How do you find the first instance of a node by element name, assuming there's more than one? – raffian Oct 11 '12 at 02:21
`xml.'**'.find { it.name() == 'elementName' }` And that will be the first one. – AbuNassar Jul 06 '16 at 13:58

4 Answers4