Reputation: 93
I'm trying to use XPath expressions using Python's ElementTree. I'm having difficulty extracting parent elements whos chilrens has a specific attribute value. These are not in the examples here, and I also tried this, this doesn't work.
My XML doc looks as follows:
<Transactions>
<Transaction>
<Project code="abc">
<Description>blah blah</Description>
<StartDate>2014-10-02</StartDate>
</Project>
<Quantity>100</Quantity>
<Price>
<Currency code="EUR" />
<Value>100</Value>
</Price>
</Transaction>
<Transaction>
<Project code="def">
<Description>something else</Description>
<StartDate>2014-10-12</StartDate>
</Project>
<Quantity>4</Quantity>
<Price>
<Currency code="EUR" />
<Value>2</Value>
</Price>
</Transaction>
<Transaction>
<Project code="abc">
<Description>blah blah</Description>
<StartDate>2014-11-02</StartDate>
</Project>
<Quantity>1</Quantity>
<Price>
<Currency code="EUR" />
<Value>123</Value>
</Price>
</Transaction></Transactions>
I'm trying to select all Transaction elements whos Project has an code of "abc".
I defined my root as follows:
transactions = ET.parse('../doc.xml').getroot();
This works (which returns all Transaction elements which have a Project as a child):
transactions.findall("Transaction[Project]")
This also works (which returns all Project elements which have the code of "abc"):
transactions.findall(".//Project[@code='abc']")
However I'm kind of lost on how to combine these (to get the Transaction elements whos child Project has a specific code). This doesn't work:
transactions.findall("Transaction[Project[@code='abc']]")
Nor this (which is discussed here):
transactions.findall("Transaction[Project/@code='abc']")
I have already spend more than 4 hours trying to solve this problem :(. If anyone is able to answer this question for me that would be very helpful!
Kind regards,
Diederik
Upvotes: 2
Views: 1619
Reputation: 473863
If you are using (or if you would switch to) lxml.etree
and would use the .xpath()
method, then your expression would work as is:
transactions.xpath("Transaction[Project[@code='abc']]")
Upvotes: 2