Question

我有以下类型的数据以xml格式返回给我（返回了很多房间;这是我回来的数据的一个例子）：

<?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>

如果nodeType == node.TEXT_NODE，我似乎能够访问数据（所以我可以看到我有1.306房间）。此外，我似乎能够访问nodeName 链接，但我真的需要知道该房间是否在我可接受的建筑物之一，所以我需要能够到达其余部分看看yyyyyyyyy。有人可以建议吗？

好的，@ vezult，正如你的建议，这是我最终想到的（工作代码！）使用ElementTree。这可能不是最pythonic（或ElementTree-ic？）这样做的方式，但似乎有效。我很高兴能够访问我的xml的每一部分的.tag，.attrib和.text。我欢迎任何有关如何使其变得更好的建议。

# 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

Answer 1

您没有提供您正在使用的库的指示，因此我假设您使用的是标准python ElementTree模块。在这种情况下，请执行以下操作：

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']

如何在python中的xml节点中访问链接数据？

1 个答案: