CODEWITHSUNDEEP
pythonxmlnodes
HelenM
HelenM

Reputation: 961

How do I access the link data in an xml node in python?

I have the following kind of data returned to me in xml format (many rooms are returned; this is one example of the data I get back):

<?xml version="1.0" encoding="UTF-8"?>
<rooms>
    <total-results>1</total-results>
    <items-per-page>1</items-per-page>
    <start-index>0</start-index>
    <room>
        <id>xxxxxxxx</id>
        <etag>5</etag>
        <link rel="http://schemas.com.mysite.building" title="building" href="https://mysite.me.myschool.edu:8443/ess/scheduleapi/v1/buildings/yyyyyyyyy"/>
        <name>1.306</name>
        <status>active</status>
        <link rel="self" title="self" href="https://mysite.me.myschool.edu:8443/ess/scheduleapi/v1/rooms/aaaaaaaaa">
    </room>
</rooms>

If nodeType == node.TEXT_NODE, I seem to be able to access the data (so I can see that I have room 1.306). Also, I seem to be able to access the nodeName link, but I really need to know if that room is in one of my acceptable buildings, so I need to be able to get to the rest of that line to look at the yyyyyyyyy. Can someone please advise?

OK, @vezult, here is what I finally came up with (working code!) using ElementTree, as you suggested. This is probably not the most pythonic (or ElementTree-ic?) way of doing this, but it seems to work. I'm thrilled to have access now to .tag, .attrib, and .text of every piece of my xml. I welcome any advice on how to make it better.

# We start out knowing our room name and our building id.  However, the same room can exist in many buildings.
# Examine the rooms we've received and get the id of the one with our name that is also in our building.

# Query the API for a list of rooms, getting u back.

request = build_request(resourceUrl)
u = urllib2.urlopen(request.to_url())
mydata = u.read()

root = ElementTree.fromstring(mydata)
print 'tree root', root.tag, root.attrib, root.text
for child in root:
    if child.tag == 'room':   
        for child2 in child:
            # the id tag comes before the name tag, so hold on to it
            if child2.tag == "id":
                hold_id = child2.text
            # the building link comes before the room name, so hold on to it
            if child2.tag == 'link':                            # if this is a link
                if "building" in child2.attrib['href']:         # and it's a building link
                    hold_link_data = child2.attrib['href']
            if child2.tag == 'name':
                if (out_bldg in hold_link_data and  # the building link we're looking at has our building in it  
                    (in_rm == child2.text)):        # and this room name is our room name
                    out_rm = hold_id
                    break  # get out of for-loop

Upvotes: 0

Views: 966

Answers (1)

vezult
vezult

Reputation: 5243

You provide no indication of what library you are using, so I'm assuming you are using the standard python ElementTree module. In that case, do the following:

from xml.etree import ElementTree

tree = ElementTree.fromstring("""<?xml version="1.0" encoding="UTF-8"?>
<rooms>
    <total-results>1</total-results>
    <items-per-page>1</items-per-page>
    <start-index>0</start-index>
    <room>
        <id>xxxxxxxx</id>
        <etag>5</etag>
        <link rel="http://schemas.com.mysite.building" title="building" href="https://mysite.me.myschool.edu:8443/ess/scheduleapi/v1/buildings/yyyyyyyyy" />
        <name>1.306</name>
        <status>active</status>
        <link rel="self" title="self" href="https://mysite.me.myschool.edu:8443/ess/scheduleapi/v1/rooms/aaaaaaaaa" />
    </room>
</rooms>
""")

# Select the first link element in the example XML
for node in tree.findall('./room/link[@title="building"]'):
    # the 'attrib' attribute is a dictionary containing the node attributes
    print node.attrib['href']

Upvotes: 3

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