[
{
"name": "Basic",
"id": "home",
"childrens": [
{
"name": "Dashboard",
"viewtype": "custom",
"view": "dashboard.html",
"childrens": []
},
{
"name": "DeviceInfo",
"href": "WSettings",
"childrens": [
{
"name": "DeviceInfo Form",
"childrens": [
{
"name": "DeviceInfo Form1",
"viewtype": "xml",
"view": "dinfo",
"childrens": []
},
{
"name": "DeviceInfo Form2",
"viewtype": "xml",
"view": "complexjson",
"childrens": []
}
]
},
{
"name": "DeviceInfo Table",
"childrens": [
{
"name": "DeviceInfo Table1",
"viewtype": "xml",
"view": "dinfotable",
"childrens": []
},
{
"name": "DeviceInfo Table2",
"viewtype": "xml",
"view": "jsontable",
"childrens": []
}
]
}
]
},
{
"name": "Hybrid",
"childrens": [
{
"name": "Table-Form",
"viewtype": "xml",
"view": "hybrid",
"childrens": []
}
]
}
]
},
{
"name": "Advanced",
"id": "profile",
"childrens": []
}
]
想要打印从根到叶的所有路径（一个带有空“子代”的路径）。例如Basic.DeviceInfo.DeviceInfo Form.DeviceInfo Form1
一切正常，直到DeviceInfo Form2
当涉及到DeviceInfo表时，就会出现DeviceInfo表单-> Basic.DeviceInfo.DeviceInfo Form.DeviceInfo Table.DeviceInfo Table1。
这不应该发生。相反，我需要Basic.DeviceInfo.DeviceInfo Table.DeviceInfo Table1。
我的代码哪里出问题了。有什么办法吗？
def walk(list1, path = ""):
for dic in list1:
#print('about to walk', dic['name'], 'passing path -->', path)
if(len(dic['childrens']) == 0):
print('leaf --->', path+dic['name']+'.')
else:
path = path+dic['name']+'.'
#passing parent name to childreni
walk(dic['childrens'], path)
最佳答案
您在else子句中设置path = path +dic['name']+'.'。一旦walk()函数完成遍历DeviceInfoForm的“子级”，它将尝试遍历DeviceInfoTable。但是，您的函数已经将路径设置为Basic.DeviceInfo.DeviceInfoForm.
您需要重新组织功能，以免在else:语句中设置路径。也许
else:
walk(dic['childrens'], path+dic['name']+'.')
这样，您会将正确的路径传递给函数，但未显式设置它。