File search in python

Question

I need a file in a dir by it's name

I achieved this with

import glob, os
os.chdir("../path")

def search_file(file_to_search):
    for file in glob.glob(file_to_search):
        return file

in another function I am using

with open("./file.csv",'rt')as f:
        lines = list(csv.reader(f))
        for line in lines:
            var = line[1]
            path = search_file(var)

Before I add the search_file function in the second process the file.csv is found. After it is not found anymore. I assume it has to do with the os import?

a work around was this

def find_file(file):
    for root, dirs, files in os.walk('./path'):
        if file in files:
            return os.path.join(root, file)

Does anyone has a solid explanation why with open() would clash with os, if that is the case?

note: the indentation might be wrong, assume the code is running, I got it from functions

Update:

I feel I need to give some more clarity:

This is what I achieve so far with what I have:

1. Function1 to read a cell from a csv file and if the name in the cell matches the name of a file in a directory, return fill path
2. Function2 to take parameter(Function1 return ) give the path to json.load(), give the load to json.dump, assign it to key(that expects json format)
3. Function processes.

This might make more sense.

Answer 1

I tried your code as below:

file.csv has 2 lines as below; file1,file1.txt file2,file2.txt

I have two txt files in path; file1.txt, file2.txt

and I tried your code it worked.

Except in your original code you have for file in glob.glob(file_file_to_search): changed to for file in glob.glob(file_to_search):

import glob, os
    os.chdir("../path")
    
def search_file(file_to_search):
    for file in glob.glob(file_to_search):
        return file

with open("file.csv",'rt')as f:
        lines = list(csv.reader(f))
        for line in lines:
            var = line[1]
            path = search_file(var)
            print(path)

Here is the ouput:

>>> with open("file.csv",'rt')as f:
...         lines = list(csv.reader(f))
...         for line in lines:
...             var = line[1]
...             path = search_file(var)
...             print(path)
...
file1.txt
file2.txt

Answer 2

Let`s say you have file.csv in your main folder and your path on Desktop. Still your code works.

import glob, os

file_to_search = "Desktop/path"

def search_file(file_to_search):
    for file in glob.glob(file_to_search):
        return file

with open("file.csv",'rt')as f:
        lines = list(csv.reader(f))
        for line in lines:
            var = line[1]
            var = file_to_search + "/" + var
            path = search_file(var)
            print(path)

Here is code on terminal:

>>> import glob, os
>>>
>>> file_to_search = "Desktop/path"
>>>
>>> def search_file(file_to_search):
...     for file in glob.glob(file_to_search):
...             return file
...
>>> with open("file.csv",'rt')as f:
...             lines = list(csv.reader(f))
...             for line in lines:
...                     var = line[1]
...                     var = file_to_search + "/" + var
...                     path = search_file(var)
...                     print(path)
...
Desktop/path/file1.txt
Desktop/path/file2.txt

Answer 3

Try this. It should search for a file, and open it if it is found.

import os
path = os.listdir('./')
newfile = input('Enter the file name: ')
for found in path:
    if str(found) == newfile:
        print(f'{str(found)} matches your search.')
        os.system(f'open {found}')
        break
    else:
        print(f'{str(found)} does not match your search.')

Answer 4

if you want get the full path of a file in a folder;

files = glob.glob("./hello/*") 

or search for a pattern; e.g., files ending with csv.

files = glob.glob("./hello/*.csv")

First check if your file is in the list and find the index of the file you are searching for you can get help from this link and if it is in the list.

the_file = files[the_index_of_the_file]

the_file has the full path.

the_file.split("/")[-1] #will give you the file name.