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" />
?我找到函数tree.find()
,但此函数的参数似乎是元素名称。
我需要根据属性name = "VHDL_BUILD_Passthrough" AND classname="TestOne"
找到节点。
答案 0 :(得分:20)
这取决于您使用的是哪个版本。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described in the docs,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),则无法使用该方便,需要自行过滤。
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']
答案 1 :(得分:0)
您必须遍历所拥有的<testcase />
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)