我使用os.listdir()，这段代码可以遍历系统目录结构，然后生成一个json文件给easyui的django前端调用display tree UI:
def CreateDirTree(path,file_json):
global id_num
id_num=0
def createDict(path):
global id_num
tree_list=[]
pathList = os.listdir(path)
for i,item in enumerate(pathList):
id_num+=1
children_map={}
children_map['id']=id_num
children_map['text']=item
if os.path.isdir(os.path.join(path,item)):
path = os.path.join(path,item)
print "directory:",path
tmp_listdir=os.listdir(path)
if len(tmp_listdir) != 0:
children_map['state']='closed'
children_map['children'] = createDict(path)
tree_list.append(children_map)
path = '\\'.join(path.split('\\')[:-1])
else:
tree_list.append(children_map)
print children_map
return tree_list
tree_list=createDict(path)
fjson = json.dumps(tree_list,ensure_ascii=False,sort_keys=False)
with open(file_json,'w') as lf:
lf.write(fjson)
##end
CreateDirTree("D:\\all_source\\somesite","123.json")
运行结果为:
directory: D:\all_source\somesite\a
file is: b.txt
file is: oo.txt
file is: a.txt
directory: D:\all_source\somesite\b
file is: 123.txt
file is: b.txt
directory: D:\all_source\somesite\c
file is: yyt.txt
directory: D:\all_source\somesite\c\z
file is: c.txt
file is: e.txt
但是我要的效果是:
directory: D:\all_source\somesite\a
a subdir files: 123.txt
a subdir files: abc.txt
directory: D:\all_source\somesite\b
directory: D:\all_source\somesite\c
directory: D:\all_source\somesite\c\z
file is: a.txt
file is: b.txt
file is: c.txt
file is: e.txt
先按目录排序再按文件排序，怎么办？请帮助我，谢谢!
最佳答案
为目录创建一个列表，为文件创建另一个列表，并迭代它们的连接。根据您的代码进行调整:
(...)
# List of directories only
dirlist = [x for x in os.listdir(path) if os.path.isdir(os.path.join(path, x))]
# List of files only
filelist = [x for x in os.listdir(path) if not os.path.isdir(os.path.join(path, x))]
for i,item in enumerate(dir_list + filelist):
(...)
这将从目录及其子目录开始，然后移动到文件
关于python - 如何使用 os.listdir 按第一个目录排序，然后在 python 中使用文件？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/29922801/