我的python脚本读取XML文件,以提供文件夹结构。
我的XML文件:
<?xml version="1.0" encoding="utf-8"?>
<folderstructure>
<folder name="Fail">
<folder name="Cam 1">
<folder name="Mod1">
<folder name="2019-04-09" />
</folder>
</folder>
<folder name="Cam 2">
<folder name="Mod1">
<folder name="2019-04-09" />
</folder>
</folder>
</folder>
<folder name="Pass">
<folder name="Cam 1">
<folder name="Mod1">
<folder name="2019-04-09" />
</folder>
</folder>
<folder name="Cam 2">
<folder name="Mod1">
<folder name="2019-04-09" />
</folder>
</folder>
</folder>
</folderstructure>
我编写了以下脚本,并参考了为所有叶节点获取路径(从根节点)(我之前的问题):
def walk(e, runningPath='', flag = 1):
name = e.attrib['name']
if len(e)>0:
runningPath += '/' + name
children = [walk(c, runningPath, 0) for c in e if ((e.tag == 'folderstructure' and flag==1) or (e.tag=='folder' and flag == 0))]
print(children)
return {'name': name, 'children': children} if children else {'name': name, 'path': runningPath + '/' + name}
但是上面的脚本产生了'None'作为输出。
我想要的输出是:
{'children': [{'children': [{'children': [{'children': [{'name': '2019-04-09',
'path': '/Fail/Cam '
'1/Mod1/2019-04-09'}],
'name': 'Mod1'}],
'name': 'Cam 1'},
{'children': [{'children': [{'name': '2019-04-09',
'path': '/Fail/Cam '
'2/Mod1/2019-04-09'}],
'name': 'Mod1'}],
'name': 'Cam 2'}],
'name': 'Fail'},
{'children': [{'children': [{'children': [{'name': '2019-04-09',
'path': '/Pass/Cam '
'1/Mod1/2019-04-09'}],
'name': 'Mod1'}],
'name': 'Cam 1'},
{'children': [{'children': [{'name': '2019-04-09',
'path': '/Pass/Cam '
'2/Mod1/2019-04-09'}],
'name': 'Mod1'}],
'name': 'Cam 2'}],
'name': 'Pass'}]
}
我该如何解决呢?
如果发生某些异常,您的函数将返回None。使用块try: except
可以捕获任何异常,因此您无法面对问题的原因,请尝试从代码中删除此块以查看问题,或者捕获更具体的异常。当我看到'folderstructure'
没有name
,你可以通过添加解决这个问题
<folderstructure name='some name'>
在你的XML或设置默认名称为您的root
元素。下面的代码似乎正常工作:
def walk(e, runningPath='', flag = 1):
try:
name = e.attrib['name']
except KeyError:
name = 'root'
if len(e)>0:
runningPath += '/' + name
children = [walk(c, runningPath, 0) for c in e if ((e.tag == 'folderstructure' and flag==1) or (e.tag=='folder' and flag == 0))]
print(children)
return {'name': name, 'children': children} if children else {'name': name, 'path': runningPath + '/' + name}
改变我使用try-catch的方式确实很有帮助。您的解决方案很优雅!我将其扩展到我的用例,而无需添加
name
属性!谢谢!!