我有300个XML文件,每个文件中都有一个路径(参见代码),我想用Python创建这个Paths的列表(.CSV)。
<da:AdminData>
<da:Datax />
<da:DataID>223</da:DataID>
<da:Date>2013-08-19</da:Date>
<da:Time>13:27:25</da:Time>
<da:Modification>2013-08-19</da:Modification>
<da:ModificationTime>13:27:25</da:ModificationTime>
**<da:Path>D:\08\06\xxx-aaa_20130806_111339.dat</da:Path>**
<da:ID>xxx-5225-fff</da:ID>
我编写了以下代码,但不适用于子目录
import os, glob, re, time, shutil
xmlpath = r'D:'
outfilename = "result.csv"
list = glob.glob(os.path.join(xmlpath,'*.xml'))
output = ""
for file in list :
fh = open(file)
text = fh.read()
pattern = "<da:Path>(.*)</da:Path>"
pattern = re.compile(pattern);
a = pattern.search(text)
if a:
output += '\n' + a.group(1)
logfile = open(outfile, "w")
logfile.write(output)
logfile.close()
答案 0 :(得分:0)
以递归方式表示,最好使用os.walk
和fnmatch.fnmatch
的组合。例如:
import os
import fnmatch
def recursive_glob(rootdir, pattern):
matching_files = []
for d, _, fnames in os.walk(rootdir):
matching_files.extend(
os.path.join(d, fname) for fname in fnames
if fnmatch.fnmatch(fname, pattern)
)
return matching_files
xmlfiles = recursive_glob(r"D:\", "*.xml")