10

我有以下 XML。

<?xml version="1.0" encoding="UTF-8"?>
<testsuites tests="10" failures="0" disabled="0" errors="0" time="0.001" name="AllTests">
  <testsuite name="TestOne" tests="5" failures="0" disabled="0" errors="0" time="0.001">
    <testcase name="DefaultConstructor" status="run" time="0" classname="TestOne" />
    <testcase name="DefaultDestructor" status="run" time="0" classname="TestOne" />
    <testcase name="VHDL_EMIT_Passthrough" status="run" time="0" classname="TestOne" />
    <testcase name="VHDL_BUILD_Passthrough" status="run" time="0" classname="TestOne" />
    <testcase name="VHDL_SIMULATE_Passthrough" status="run" time="0.001" classname="TestOne" />
</testsuite>
</testsuites>

问:如何找到节点<testcase name="VHDL_BUILD_Passthrough" status="run" time="0" classname="TestOne" />?我找到了 function tree.find(),但这个函数的参数似乎是元素名称。

我需要根据属性找到节点:name = "VHDL_BUILD_Passthrough" AND classname="TestOne"

4

2 回答 2

24

这取决于您使用的版本。如果您有 ElementTree 1.3+(包括在 Python 2.7 标准库中),您可以使用基本的 xpath 表达式,如文档中所述,例如[@attrib='value']

x = ElmentTree(file='testdata.xml')
cases = x.findall(".//testcase[@name='VHDL_BUILD_Passthrough'][@classname='TestOne']")

不幸的是,如果您使用的是早期版本的 ElementTree(1.2,包含在 python 2.5 和 2.6 的标准库中),您将无法使用这种便利,需要自己过滤。

x = ElmentTree(file='testdata.xml')
allcases = x12.findall(".//testcase")
cases = [c for c in allcases if c.get('classname') == 'TestOne' and c.get('name') == 'VHDL_BUILD_Passthrough']
于 2011-01-26T19:57:15.517 回答
0

您必须遍历<testcase />您拥有的元素,如下所示:

from xml.etree import cElementTree as ET

# assume xmlstr contains the xml string as above
# (after being fixed and validated)
testsuites = ET.fromstring(xmlstr)
testsuite = testsuites.find('testsuite')
for testcase in testsuite.findall('testcase'):
    if testcase.get('name') == 'VHDL_BUILD_Passthrough':
        # do what you will with `testcase`, now it is the element
        # with the sought-after attribute
        print repr(testcase)
于 2011-01-26T19:50:15.437 回答