获取特定 XML 个元素的值并将它们添加到 Python 中的数组

Question

以下面的 XML 为例，我试图获取所有 <d:LayerXml> 标签的内容并将它们添加到数组中。要解析 XML 我正在使用 ElementTree。

我首先尝试使用那里的名称访问 XML 元素，但这失败了，因为显然没有名为 'entry' -

的元素

root = ET.fromstring(r.text)
for child in root:
    if child.tag == entry':
        print child.attirb

在打印出所有子标签 (print child.tag) 后，我注意到每个子标签都带有 roor 元素中提供的 xmlns 后缀。例如，'entry' 实际上是 '{http://www.w3.org/2005/Atom}'。

所以接下来我尝试使用该后缀访问元素，但由于语法错误而失败。

root = ET.fromstring(r.text)
for child in root:
    if child.tag == '{http://www.w3.org/2005/Atom}entry':
        layerXML = child.{http://www.w3.org/2005/Atom}content
# Also tried - layerXML = child.'{http://www.w3.org/2005/Atom}content'
        print layerXML

所以给定以下 XML 示例，我如何将所有 <d:LayerXml> 元素的内容添加到数组中。澄清一下，在这种情况下，数组将包含 I want this 和 I want this, too.

<?xml version="1.0" encoding="utf-8"?>
<feed xml:base="https://tablestore.somewhere.com/" xmlns="http://www.w3.org/2005/Atom" xmlns:d="http://schemas.microsoft.com/ado/2007/08/dataservices" xmlns:m="http://schemas.microsoft.com/ado/2007/08/dataservices/metadata" xmlns:georss="http://www.georss.org/georss" xmlns:gml="http://www.opengis.net/gml">
    <id>https://tablestore.somewhere.com/TableName</id>
    <title type="text">TableName</title>
    <updated>2017-03-02T12:01:04Z</updated>
    <link rel="self" title="TableName" href="TableName" />
    <entry m:etag="W/&quot;datetime'2017-03-02T11%3A46%3A37.1271167Z'&quot;">
        <id>https://tablestore.somewhere.com/TableName(PartitionKey='PartitonKey',RowKey='layer1-tileMatrixSet')</id>
        <category term="tablestore.TableName" scheme="http://schemas.microsoft.com/ado/2007/08/dataservices/scheme" />
        <link rel="edit" title="TableName" href="TableName(PartitionKey='PartitonKey',RowKey='layer1-tileMatrixSet')" />
        <title />
        <updated>2017-03-02T12:01:04Z</updated>
        <author>
            <name />
        </author>
        <content type="application/xml">
            <m:properties>
                <d:PartitionKey>PartitonKey</d:PartitionKey>
                <d:RowKey>RowKey</d:RowKey>
                <d:Timestamp m:type="Edm.DateTime">2017-03-02T11:46:37.1271167Z</d:Timestamp>
                <d:AuthType>basic</d:AuthType>
                <d:Credentials>CREDENTIALS1</d:Credentials>
                <d:Layer>layer1</d:Layer>
                <d:LayerXml>I want this</d:LayerXml>
                <d:Service>https://www.google.co.uk</d:Service>
                <d:TileMatrixSet>tileMatrixSet</d:TileMatrixSet>
            </m:properties>
        </content>
    </entry>
    <entry m:etag="W/&quot;datetime'2017-03-02T11%3A46%3A37.1271167Z'&quot;">
        <id>https://tablestore.somewhere.com/TableName(PartitionKey='PartitonKey',RowKey='layer2-tileMatrixSet')</id>
        <category term="tablestore.TableName" scheme="http://schemas.microsoft.com/ado/2007/08/dataservices/scheme" />
        <link rel="edit" title="TableName" href="TableName(PartitionKey='PartitonKey',RowKey='layer2-tileMatrixSet')" />
        <title />
        <updated>2017-03-02T12:01:04Z</updated>
        <author>
            <name />
        </author>
        <content type="application/xml">
            <m:properties>
                <d:PartitionKey>PartitonKey</d:PartitionKey>
                <d:RowKey>RowKey</d:RowKey>
                <d:Timestamp m:type="Edm.DateTime">2017-03-02T11:46:37.1271167Z</d:Timestamp>
                <d:AuthType>basic</d:AuthType>
                <d:Credentials>CREDENTIALS1</d:Credentials>
                <d:Layer>layer2</d:Layer>
                <d:LayerXml>I want this, too</d:LayerXml>
                <d:Service>https://www.google.co.uk</d:Service>
                <d:TileMatrixSet>tileMatrixSet</d:TileMatrixSet>
            </m:properties>
        </content>
    </entry>
</feed>

Answer 1

你还没说是哪个语法错误，下面给出了我想要的结果：

from xml.etree import ElementTree as ET
xml = '''<?xml version="1.0" encoding="utf-8"?>
<feed xml:base="https://tablestore.somewhere.com/" xmlns="http://www.w3.org/2005/Atom" xmlns:d="http://schemas.microsoft.com/ado/2007/08/dataservices" xmlns:m="http://schemas.microsoft.com/ado/2007/08/dataservices/metadata" xmlns:georss="http://www.georss.org/georss" xmlns:gml="http://www.opengis.net/gml">
    <id>https://tablestore.somewhere.com/TableName</id>
    <title type="text">TableName</title>
    <updated>2017-03-02T12:01:04Z</updated>
    <link rel="self" title="TableName" href="TableName" />
    <entry m:etag="W/&quot;datetime'2017-03-02T11%3A46%3A37.1271167Z'&quot;">
        <id>https://tablestore.somewhere.com/TableName(PartitionKey='PartitonKey',RowKey='layer1-tileMatrixSet')</id>
        <category term="tablestore.TableName" scheme="http://schemas.microsoft.com/ado/2007/08/dataservices/scheme" />
        <link rel="edit" title="TableName" href="TableName(PartitionKey='PartitonKey',RowKey='layer1-tileMatrixSet')" />
        <title />
        <updated>2017-03-02T12:01:04Z</updated>
        <author>
            <name />
        </author>
        <content type="application/xml">
            <m:properties>
                <d:PartitionKey>PartitonKey</d:PartitionKey>
                <d:RowKey>RowKey</d:RowKey>
                <d:Timestamp m:type="Edm.DateTime">2017-03-02T11:46:37.1271167Z</d:Timestamp>
                <d:AuthType>basic</d:AuthType>
                <d:Credentials>CREDENTIALS1</d:Credentials>
                <d:Layer>layer1</d:Layer>
                <d:LayerXml>I want this</d:LayerXml>
                <d:Service>https://www.google.co.uk</d:Service>
                <d:TileMatrixSet>tileMatrixSet</d:TileMatrixSet>
            </m:properties>
        </content>
    </entry>
    <entry m:etag="W/&quot;datetime'2017-03-02T11%3A46%3A37.1271167Z'&quot;">
        <id>https://tablestore.somewhere.com/TableName(PartitionKey='PartitonKey',RowKey='layer2-tileMatrixSet')</id>
        <category term="tablestore.TableName" scheme="http://schemas.microsoft.com/ado/2007/08/dataservices/scheme" />
        <link rel="edit" title="TableName" href="TableName(PartitionKey='PartitonKey',RowKey='layer2-tileMatrixSet')" />
        <title />
        <updated>2017-03-02T12:01:04Z</updated>
        <author>
            <name />
        </author>
        <content type="application/xml">
            <m:properties>
                <d:PartitionKey>PartitonKey</d:PartitionKey>
                <d:RowKey>RowKey</d:RowKey>
                <d:Timestamp m:type="Edm.DateTime">2017-03-02T11:46:37.1271167Z</d:Timestamp>
                <d:AuthType>basic</d:AuthType>
                <d:Credentials>CREDENTIALS1</d:Credentials>
                <d:Layer>layer2</d:Layer>
                <d:LayerXml>I want this, too</d:LayerXml>
                <d:Service>https://www.google.co.uk</d:Service>
                <d:TileMatrixSet>tileMatrixSet</d:TileMatrixSet>
            </m:properties>
        </content>
    </entry>
</feed>
'''
feed = ET.fromstring(xml)
values = [value.text for value in feed.findall('{http://www.w3.org/2005/Atom}entry/{http://www.w3.org/2005/Atom}content/{http://schemas.microsoft.com/ado/2007/08/dataservices/metadata}properties/{http://schemas.microsoft.com/ado/2007/08/dataservices}LayerXml')]
print(values)

其实好像也可以用

values = [value.text for value in feed.findall('.//{http://schemas.microsoft.com/ado/2007/08/dataservices}LayerXml')]

或

values = [value.text for value in feed.findall('.//d:LayerXml', { 'd' : 'http://schemas.microsoft.com/ado/2007/08/dataservices' })]

如果您不想列出完整路径。

获取特定 XML 个元素的值并将它们添加到 Python 中的数组

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