Question

我需要以JSON格式获得一种过滤的目录/文件结构。

具体来说，我需要只包含包含给定字符串的文件，并且只包含包含此类文件的目录（本身或其某些后代）。

此代码：

import os
import json

def path_to_dict(path):
    d = {'name': os.path.basename(path)}
    if os.path.isdir(path):
        d['type'] = "directory"
        d['children'] = [path_to_dict(os.path.join(path,x)) for x in os.listdir\
(path)]
    else:
        d['type'] = "file"
    return d

print json.dumps(path_to_dict('.'), indent=2)

以我想要的格式从当前目录开始，为我提供了所有目录和文件的精美JSON树：

{
    "type": "directory",
    "name": ".",
    "children": [
    {
      "type": "file", 
      "name": "attribute_container.c"
    }, 
    {
      "type": "file", 
      "name": "node.c"
    }, 
    {
      "type": "directory", 
      "name": "power", 
      "children": [
        {
          "type": "file", 
          "name": "clock_ops.c"
        }, 
        {
          "type": "file", 
          "name": "common.c"
        }, 
        {
          "type": "file", 
          "name": "domain.c"
        }, 
        {
          "type": "file", 
          "name": "domain_governor.c"
        }, 
        {
          "type": "file", 
          "name": "generic_ops.c"
        }, 
        {
          "type": "file", 
          "name": "wakeup.c"
        }
      ]
    }, 
    {
      "type": "directory", 
      "name": "regmap", 
      "children": [
        {
          "type": "file", 
          "name": "internal.h"
        }, 
        {
          "type": "file", 
          "name": "Kconfig"
        }, 
        {
          "type": "file", 
          "name": "Makefile"
        }, 
        {
          "type": "file", 
          "name": "regcache-flat.c"
        }, 
        {
          "type": "file", 
          "name": "regmap-spmi.c"
        }, 
        {
          "type": "file", 
          "name": "regmap.c"
        }
      ]
    }, 
    {
      "type": "file", 
      "name": "soc.c"
    }, 
    {
      "type": "file", 
      "name": "syscore.c"
    }, 
    {
      "type": "file", 
      "name": "topology.c"
    }, 
    {
      "type": "file", 
      "name": "transport_class.c"
    }   ] }

但是，我只需要包含给定字符串的文件。此外，只有包含此类文件或文件或其某些后代的文件夹才包含此类文件。（可以这么说，我需要一种＆＃34;修剪＆＃34;）

我知道在文件中找到字符串的解决方案：

my_file = ...
my_string = ...
infile = open(my_file,"r")

numlines = 0
found = 0
for line in infile:
    numlines += 1
    found += line.count(my_string)
infile.close()

print "%s was found %i times in %i lines", %string, %found, %numlines

但我很难将其整合到问题顶部的代码中。

我感谢任何提示或建议。

Answer 1

我不想使用os.walk()重写您的代码。我会对您进行一些小改动。

密钥是使用None作为修剪文件的sentinel值，使用空children列表来修剪目录。该实现并没有很好地编写，但它向您展示了如何使用测试的核心。

import os
import json

def check_in_file(my_file,my_string):
    with open(my_file) as f:
        try:
            return my_string in f.read()
        except:
            return False

def path_to_dict(path, my_string=None):
    d = {'name': os.path.basename(path)}
    if os.path.isdir(path):
        d['type'] = "directory"
        d['children'] = []
        paths = [os.path.join(path,x) for x in os.listdir(path)]
        #Just the children that contains at least a valid file
        for p in paths:
            c = path_to_dict(p, my_string)
            if c is not None:
                d['children'].append(c)
        if not d['children']:
            return None
    else:
        if my_string is not None and not check_in_file(path,my_string):
            return None
        d['type'] = "file"
    return d

print(json.dumps(path_to_dict('.',), indent=2))
print(json.dumps(path_to_dict('.','kkkkk'), indent=2))

获取文件夹和文件的JSON树（但仅包含包含给定字符串的文件）

1 个答案: