Can i dereference an lxml.etree.AncestorsIterator?

Question

I'm using lxml to manipulate a dbschema expressed in an xml file. It looks something like this:

<Tables>
<Table name = "table1">
<Columns>
<Column name="COL1">...</Column>
<Column name="COL2">...
    <References>
    <Reference>TABLENAME</Reference>
    </References>
</Column>
</Table>
...
</Tables>

Currently I want to look at the references, and get the table and column names for those references. The following works:

refiter = mytree.iter("Reference")
for r in refiter:
   nameiter =r.iterancestors("Table")
   for n in nameiter:
       tablename = .get("name")

I don't like this solution, because I know that my nameiter can only iterate over a single element -- it only has one parent "Table". It seems that in python I can only use an iterator in a loop. But I find it a bit silly. I know I have only one ancestor "Table". Can I dereference the iterator directly somehow? Or is there an alternative method to getting this information that's more suitable?

Answer 1

You can do it using an xpath to get both the ancestors you want

x = """<?xml version="1.0" encoding="utf-8"?>
<Tables>
<Table name = "table1">
<Columns>
<Column name="COL1">...</Column>
<Column name="COL2">...
    <References>
    <Reference>TABLENAME</Reference>
    </References>
</Column>
</Columns>
</Table>
<Table name = "table2">
<Columns>
<Column name="COL2">...</Column>
<Column name="COL3">...
    <References>
    <Reference>TABLENAME</Reference>
    </References>
</Column>
</Columns>
</Table>
</Tables>"""


import lxml.etree  as et

xml = et.fromstring(x)

refs = xml.iter("Reference")
print([(ref.xpath("./ancestor::Table/@name")[0], ref.xpath("./ancestor::Column/@name")[0]) for ref in refs])

Which would give you:

[('table1', 'COL2'), ('table2', 'COL3')]

Or if the Column is always the grandparent:

 [(ref.xpath("./ancestor::Table/@name")[0], ref.xpath("./../../@name")[0]) for ref in refs]

using your own logic, you can just call next on iterancetors:

refs = xml.iter("Reference")


for r in refs:
   print(next(r.iterancestors("Table")).get("name"))
   print(next(r.iterancestors("Column")).get("name"))

Which would give you:

table1
COL2
table2
COL3

Answer 2

As you are interested in only the first result of the iterator, you can use the next method to get the first element, and avoid the unclear/unnecessary for loop.

xml_string = """
<Tables>
<Table name = "table1">
<Columns>
<Column name="COL1">...</Column>
<Column name="COL2">...
    <References>
    <Reference>TABLENAME</Reference>
    </References>
</Column>
</Columns>
</Table>
<Table name = "table2">
<Columns>
<Column name="COL2">...</Column>
<Column name="COL3">...
    <References>
    <Reference>TABLENAME</Reference>
    </References>
</Column>
</Columns>
</Table>
</Tables>"""


import lxml.etree as ETree

root = ETree.fromstring(bytes(xml_string, 'UTF-8'))

refiter = root.iter('Reference')
for r in refiter:
    nameiter = r.iterancestors('Table')
    name = next(nameiter).get('name')
    print(name)

If you wanted to access results by index, you can generate a list from the iterator first.

tables = list(r.iterancestors('Table'))
print(tables[0].get('name'))