解析根文件夹及其子文件夹中的xml文件

Question

我正在处理一个包含 27 个文件夹和每个文件夹内的各种 XML 文件的目录。

能够：

• 解析一个 XML 文件并写入一个 CSV 文件
• 遍历一个文件夹，读取并解析其中的所有XML文件

挑战：

• 在尝试将所有 XML 文件从根文件夹解析到其子文件夹时遇到问题

发送帮助并感谢您。代码如下：sn-ps

# working in one folder only
import csv
import xml.etree.ElementTree as ET
import os

## directory
path = "/Users.../y"
filenames = []

## Count the number of xml files of each folder
files = os.listdir(path)
print("\n")


xml_data_to_csv = open('/Users.../xml_extract.csv', 'w')
list_head = []
csvwriter = csv.writer(xml_data_to_csv)

# Read XML files in a folder
for filename in os.listdir(path):
    if not filename.endswith('.xml'):
        continue
    fullname = os.path.join(path,filename)
    print("\n", fullname)
    filenames.append(fullname)

# parse elements in each XML file
for filename in filenames:
    tree = ET.parse(filename)
    root = tree.getroot()

    extract_xml=[]

    ## extract child elements per xml file
    print("\n")
    for x in root.iter('Info'):
        for element in x:
            print(element.tag,element.text)
            extract_xml.append(element.text)


    ## Write list nodes to csv
    csvwriter.writerow(extract_xml)

## Close CSV file
xml_data_to_csv.close()

Answer 1

您可以通过

获取给定路径中所有 XML 文件的列表

import os

path = "main/root"
filelist = []

for root, dirs, files in os.walk(path):
    for file in files:
        if not file.endswith('.xml'):
            continue
        filelist.append(os.path.join(root, file))

for file in filelist:
    print(file)
    # or in your case parse the XML 'file'

例如：

$ tree /main/root
/main/root
├── a
│   ├── a.xml
│   ├── b.xml
│   └── c.xml
├── b
│   ├── d.xml
│   ├── e.xml
│   └── x.txt
└── c
    ├── f.xml
    └── g.xml

我们得到：

/main/root/c/g.xml
/main/root/c/f.xml
/main/root/b/e.xml
/main/root/b/d.xml
/main/root/a/c.xml
/main/root/a/b.xml
/main/root/a/a.xml

如果要对目录和文件进行排序：

for root, dirs, files in os.walk(path):
    dirs.sort()
    for file in sorted(files):
        if not file.endswith('.xml'):
            continue
        filelist.append(os.path.join(root, file))

Answer 2

你可以使用os.walk:

import os
for dir_name, dirs, files in os.walk('<root_dir>'):
    # parse files

Answer 3

您可以使用pathlib 模块来“glob” XML 文件。它将在所有子目录中搜索您提供的模式并返回已包含文件路径的 Path 对象。稍微清理一下你的脚本，你就会有

import csv
import xml.etree.ElementTree as ET
from pathlib import Path

## directory
path = Path("/Users.../y")

with open('/Users.../xml_extract.csv', 'w') as xml_data_to_csv:
    csvwriter = csv.writer(xml_data_to_csv)

    # Read XML files in a folder
    for filepath in path.glob("**/*.xml"):
        tree = ET.parse(filename)
        root = tree.getroot()

        extract_xml=[]

        ## extract child elements per xml file
        print("\n")
        for x in root.iter('Info'):
            for element in x:
                print(element.tag,element.text)
                extract_xml.append(element.text)

        ## Write list nodes to csv
        csvwriter.writerow(extract_xml)