从文件中的一行读取特定字符串

Question

I have file looking like this:

face LODRxERROR
{
  source   R/com/int/LRxAMEexception.csv

  contains R/saqf/LAWODRxERROR.ddf
  contains R/bld/LAWODRxERRORtyp.h
  contains R/bld/LAWODRxERRORtyp.hpp

  requires LAWODRxERR
}

At the moment I'm able to read a specific line and store it. But I need to be more specific. Instead of reading the whole line. I would like to read only the file name no the directory. So, instead of reading R/bld/LAWODRxERRORtyp.hpp I would like to read only LAWODRxERRORtyp.hpp

Here is my python code so far:

    with open(file) as scope:
        for line in scope:
            line = line.strip()
            if line.startswith('contains') and line.endswith('.h') or line.endswith('.hpp'):
                scopeFileList.append(line.split()[-1])

Thanks in advance

Answer 1

You can use the built-in function os.path.basename() to get only the file-name from a path:

from os.path import basename

with open(file) as scope:
    for line in scope:
        line = line.strip()
        if line.startswith('contains') and line.endswith('.h') or line.endswith('.hpp'):
            path = line.split()[-1]
            scopeFileList.append(basename(path))

Answer 2

Try this,

with open("file1.txt", "r") as f:
    data = [line.replace("\n","").split('/')[-1] for line in f.readlines() if '.' in line]

---

Output:

print(data)

['LRxAMEexception.csv',
 'LAWODRxERROR.ddf',
 'LAWODRxERRORtyp.h',
 'LAWODRxERRORtyp.hpp']

Answer 3

Try this: You can use the re.search to find the file names from a path

with open('new_file.txt') as file:
     for line in file:
             line = line.strip()
             if line.startswith('source') or line.startswith('contains'):
                    file_names = re.search('/(.+?)\/((\w+)\.\w+$\Z)', line).group(2)
                     print(file_names)

O/P:

'LRxAMEexception.csv'
'LAWODRxERROR.ddf'
'LAWODRxERRORtyp.h'
'LAWODRxERRORtyp.hpp'