Parsing xml with etree

Question

I am trying to parse an XML response from Amazon's Product Advertising API, this is the xml

<?xml version="1.0" ?>
    <ItemLookupResponse xmlns="http://webservices.amazon.com/AWSECommerceService/2010-11-01"> <OperationRequest>
        <HTTPHeaders>
            <Header Name="UserAgent" Value="TSN (Language=Python)"></Header>
        </HTTPHeaders>
        <RequestId>96ef9bc3-68a8-4bf3-a2c7-c98b8aeae00f</RequestId>
        <Arguments>
            <Argument Name="Operation" Value="ItemLookup"></Argument>
            <Argument Name="Service" Value="AWSECommerceService"></Argument>
            <Argument Name="Signature" Value="gjc4wRNum3YT82app1d06vMIDM7v44fOmZTP8Uh3LqE="></Argument><Argument Name="AssociateTag" Value="sneakick-20"></Argument>
            <Argument Name="Version" Value="2010-11-01"></Argument>
            <Argument Name="ItemId" Value="810056013349,810056013264"></Argument>
            <Argument Name="IdType" Value="UPC"></Argument>
            <Argument Name="AWSAccessKeyId" Value="AKIAIFMUMJLJOOINRVRA"></Argument>
            <Argument Name="Timestamp" Value="2012-01-03T21:26:39Z"></Argument>
            <Argument Name="ResponseGroup" Value="ItemIds"></Argument>
            <Argument Name="SearchIndex" Value="Apparel"></Argument>
        </Arguments>
       <RequestProcessingTime>0.0595830000000000</RequestProcessingTime>
      </OperationRequest>
      <Items>
          <Request>
              <IsValid>True</IsValid>
              <ItemLookupRequest>
                  <IdType>UPC</IdType>
                  <ItemId>810056013349</ItemId>
                  <ItemId>810056013264</ItemId>
                  <ResponseGroup>ItemIds</ResponseGroup>
                  <SearchIndex>Apparel</SearchIndex>
                  <VariationPage>All</VariationPage>
              </ItemLookupRequest>
          </Request>
          <Item>
              <ASIN>B000XR4K6U</ASIN>
          </Item>
          <Item>
              <ASIN>B000XR2UU8</ASIN>
          </Item>
       </Items>
    </ItemLookupResponse>

All i am interested in is the Item tags inside Items , so basically all that xml was returned by amazon in a string which i parsed like so:

from xml.etree.ElementTree import fromstring

response = "xml string returned by amazon"
parsed = fromstring(response)
items = parsed[1] # This is how i get the Items element

# These were my attempts at getting the Item element
items.find('Item')
items.findall('Item')

items being the Items element, but so far no success, it keeps returning None/Empty , im i missing something , or is there another way to go about this ?

Would be helpful if you would show your parsing code-part! – Don Question Jan 03 '12 at 22:07 — Don Question, Jan 03 '12 at 22:07

mzjn · Accepted Answer · 2012-01-04T18:01:39.110

It is a namespace issue. This works:

from xml.etree import ElementTree as ET

XML = """<?xml version="1.0" ?>
    <ItemLookupResponse xmlns="http://webservices.amazon.com/AWSECommerceService/2010-11-01"> 
      <OperationRequest>
        <HTTPHeaders>
            <Header Name="UserAgent" Value="TSN (Language=Python)"></Header>
        </HTTPHeaders>
        <RequestId>96ef9bc3-68a8-4bf3-a2c7-c98b8aeae00f</RequestId>
        <Arguments>
            <Argument Name="Operation" Value="ItemLookup"></Argument>
            <Argument Name="Service" Value="AWSECommerceService"></Argument>
            <Argument Name="Signature" Value="gjc4wRNum3YT82app1d06vMIDM7v44fOmZTP8Uh3LqE="></Argument>
            <Argument Name="AssociateTag" Value="sneakick-20"></Argument>
            <Argument Name="Version" Value="2010-11-01"></Argument>
            <Argument Name="ItemId" Value="810056013349,810056013264"></Argument>
            <Argument Name="IdType" Value="UPC"></Argument>
            <Argument Name="AWSAccessKeyId" Value="AKIAIFMUMJLJOOINRVRA"></Argument>
            <Argument Name="Timestamp" Value="2012-01-03T21:26:39Z"></Argument>
            <Argument Name="ResponseGroup" Value="ItemIds"></Argument>
            <Argument Name="SearchIndex" Value="Apparel"></Argument>
        </Arguments>
       <RequestProcessingTime>0.0595830000000000</RequestProcessingTime>
      </OperationRequest>
      <Items>
          <Request>
              <IsValid>True</IsValid>
              <ItemLookupRequest>
                  <IdType>UPC</IdType>
                  <ItemId>810056013349</ItemId>
                  <ItemId>810056013264</ItemId>
                  <ResponseGroup>ItemIds</ResponseGroup>
                  <SearchIndex>Apparel</SearchIndex>
                  <VariationPage>All</VariationPage>
              </ItemLookupRequest>
          </Request>
          <Item>
              <ASIN>B000XR4K6U</ASIN>
          </Item>
          <Item>
              <ASIN>B000XR2UU8</ASIN>
          </Item>
       </Items>
    </ItemLookupResponse>"""

NS = "{http://webservices.amazon.com/AWSECommerceService/2010-11-01}"

doc = ET.fromstring(XML)
Item_elems = doc.findall(".//" + NS + "Item")  # All Item elements in document

print Item_elems

Output:

[<Element '{http://webservices.amazon.com/AWSECommerceService/2010-11-01}Item' at 0xbf0c50>, 
<Element '{http://webservices.amazon.com/AWSECommerceService/2010-11-01}Item' at 0xbf0cd0>]

Variation closer to your own code:

NS = "{http://webservices.amazon.com/AWSECommerceService/2010-11-01}"
doc = ET.fromstring(XML)
items = doc[1]                           # Items element

first_item = items.find(NS + 'Item')     # First direct Item child
all_items =  items.findall(NS + 'Item')  # List of all direct Item children

I wish I could upvote this +10. Just because of the last code sample. Nice examples are somehow unnervingly hard to find. — sjas, Jan 28 '13 at 02:35

score 1 · Answer 2 · edited May 23 '17 at 12:24

Namespace issue.

You can put the namespace in front of all of your items as spelled out in the first answer to either this question or this question. A possibly simpler solution is to ignore the namespace with a quick hack like this:

xml_hacked_namespace = raw_xml.replace(' xmlsn=', ' xmlnamespace=')
doc = fromstring(xml_hacked_namespace)
item_list = doc.findall('.//Item')

If you find that you are doing a lot of work with xml you may also be interested in checking out lxml. It is faster and provides a few extra methods that some find nice to have.

Parsing xml with etree

2 Answers2