Python lxml XPath : preceding keyword does not give expected result

Question

i am trying to parse an xml document as follows

import re
from lxml.html.soupparser import fromstring

inString = """
<doc>

<q></q>

<p1>
    <p2 dd="ert" ji="pp">

        <p3>1</p3>
        <p3>2</p3>
        <p3>ABC</p3>
        <p3>3</p3>

     </p2>

     <p2 dd="ert" ji="pp">

        <p3>4</p3>
        <p3>5</p3>
        <p3>ABC</p3>
        <p3>6</p3>

     </p2>

</p1>
<r></r>
<p1>
    <p2 dd="ert" ji="pp">

        <p3>7</p3>
        <p3>8</p3>
        <p3>ABC</p3>
        <p3>9</p3>

     </p2>

     <p2 dd="ert" ji="pp">

        <p3>10</p3>
        <p3>11</p3>
        <p3>ABC</p3>
        <p3>12</p3>

     </p2>

</p1>
</doc>
"""
root = fromstring(inString)

nodes = root.xpath("./doc//p1/p2/p3[contains(text(),'ABC')]//preceding::p2//p3")

print " ".join([re.sub('[\s+]', ' ', para.text.encode('utf-8').strip()) for para in nodes])

so, for each <p1> tag, i want to get to <p3> tags inside <p2> . Then i only want the <p3> tags upto tag having text like ABC . however, if i run the above code, i get

1 2 ABC 3 4 5 ABC 6 7 8 ABC 9

desired output is

1 2 4 5 7 8 10 11

also, if i make this change

nodes = root.xpath("./doc//p1/p2/p3[contains(text(),'ABC')]")

i get

ABC ABC ABC ABC

so looks like the second approach is able to get all the <p3> nodes from the entire document as per the xpath, which is fine. why doesn't my first query work?

how do i get the desired output?

Answer 1

Once you've located the p3 containing ABC , you don't need to get up the tree - just go "sideways" using the preceding-sibling :

./doc//p1/p2/p3[contains(text(),'ABC')]/preceding-sibling::p3

Prints 1 2 4 5 7 8 10 11 .