PHP dirname () function

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The dirname() function returns the path to the parent directory, which includes a period (’. ’) unless the path contains a forward slash indicating the current directory.Syntax:
string dirname ($path)
Parameters:The dirname() function in PHP only takes one parameter, which is $path. This parameter specifies the path to check.Return value:returns the path to the parent directory.Errors and exceptions :
  • When specifying a path, both forward slashes, forward slashes (/) and backslashes (/) are used as a directory separator character in Windows environment, whereas in other environments it is just forward slash (/) .
  • The dirname() function operates on an input string and therefore does not know about the actual filesystem or path components such as "..".
  • Examples :
    Input: dirname ("user01 / engineerforengineer / gfg.txt") Output: user01 / engineerforengineer Input: dirname ("/ engineerforengineer / gfg.txt"); Output: / engineerforengineer
    The programs below illustrate the dirname() function:Program 1 :
    // specifying the path to the function dirname() echo dirname ( " user01 / engineerforengineer / gfg.txt " )
    ?>
    Output:
    user01 / engineerforengineer
    Program 2 :
    // specifying the path to the dirname() function echo dirname ( " / engineerforengineer / gfg.txt " );
    ?>
    Output:
    / engineerforengineer 
    Link:
    http://php.net/manual /en/function.dirname.php

    PHP dirname () function _files: Questions

    How do I list all files of a directory?

    5 answers

    How can I list all files of a directory in Python and add them to a list?

    3467

    Answer #1

    os.listdir() will get you everything that"s in a directory - files and directories.

    If you want just files, you could either filter this down using os.path:

    from os import listdir
    from os.path import isfile, join
    onlyfiles = [f for f in listdir(mypath) if isfile(join(mypath, f))]
    

    or you could use os.walk() which will yield two lists for each directory it visits - splitting into files and dirs for you. If you only want the top directory you can break the first time it yields

    from os import walk
    
    f = []
    for (dirpath, dirnames, filenames) in walk(mypath):
        f.extend(filenames)
        break
    

    or, shorter:

    from os import walk
    
    filenames = next(walk(mypath), (None, None, []))[2]  # [] if no file
    

    3467

    Answer #2

    I prefer using the glob module, as it does pattern matching and expansion.

    import glob
    print(glob.glob("/home/adam/*"))
    

    It does pattern matching intuitively

    import glob
    # All files ending with .txt
    print(glob.glob("/home/adam/*.txt")) 
    # All files ending with .txt with depth of 2 folder
    print(glob.glob("/home/adam/*/*.txt")) 
    

    It will return a list with the queried files:

    ["/home/adam/file1.txt", "/home/adam/file2.txt", .... ]
    

    3467

    Answer #3

    os.listdir() - list in the current directory

    With listdir in os module you get the files and the folders in the current dir

     import os
     arr = os.listdir()
     print(arr)
     
     >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]
    

    Looking in a directory

    arr = os.listdir("c:\files")
    

    glob from glob

    with glob you can specify a type of file to list like this

    import glob
    
    txtfiles = []
    for file in glob.glob("*.txt"):
        txtfiles.append(file)
    

    glob in a list comprehension

    mylist = [f for f in glob.glob("*.txt")]
    

    get the full path of only files in the current directory

    import os
    from os import listdir
    from os.path import isfile, join
    
    cwd = os.getcwd()
    onlyfiles = [os.path.join(cwd, f) for f in os.listdir(cwd) if 
    os.path.isfile(os.path.join(cwd, f))]
    print(onlyfiles) 
    
    ["G:\getfilesname\getfilesname.py", "G:\getfilesname\example.txt"]
    

    Getting the full path name with os.path.abspath

    You get the full path in return

     import os
     files_path = [os.path.abspath(x) for x in os.listdir()]
     print(files_path)
     
     ["F:\documentiapplications.txt", "F:\documenticollections.txt"]
    

    Walk: going through sub directories

    os.walk returns the root, the directories list and the files list, that is why I unpacked them in r, d, f in the for loop; it, then, looks for other files and directories in the subfolders of the root and so on until there are no subfolders.

    import os
    
    # Getting the current work directory (cwd)
    thisdir = os.getcwd()
    
    # r=root, d=directories, f = files
    for r, d, f in os.walk(thisdir):
        for file in f:
            if file.endswith(".docx"):
                print(os.path.join(r, file))
    

    os.listdir(): get files in the current directory (Python 2)

    In Python 2, if you want the list of the files in the current directory, you have to give the argument as "." or os.getcwd() in the os.listdir method.

     import os
     arr = os.listdir(".")
     print(arr)
     
     >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]
    

    To go up in the directory tree

    # Method 1
    x = os.listdir("..")
    
    # Method 2
    x= os.listdir("/")
    

    Get files: os.listdir() in a particular directory (Python 2 and 3)

     import os
     arr = os.listdir("F:\python")
     print(arr)
     
     >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]
    

    Get files of a particular subdirectory with os.listdir()

    import os
    
    x = os.listdir("./content")
    

    os.walk(".") - current directory

     import os
     arr = next(os.walk("."))[2]
     print(arr)
     
     >>> ["5bs_Turismo1.pdf", "5bs_Turismo1.pptx", "esperienza.txt"]
    

    next(os.walk(".")) and os.path.join("dir", "file")

     import os
     arr = []
     for d,r,f in next(os.walk("F:\_python")):
         for file in f:
             arr.append(os.path.join(r,file))
    
     for f in arr:
         print(files)
    
    >>> F:\_python\dict_class.py
    >>> F:\_python\programmi.txt
    

    next(os.walk("F:\") - get the full path - list comprehension

     [os.path.join(r,file) for r,d,f in next(os.walk("F:\_python")) for file in f]
     
     >>> ["F:\_python\dict_class.py", "F:\_python\programmi.txt"]
    

    os.walk - get full path - all files in sub dirs**

    x = [os.path.join(r,file) for r,d,f in os.walk("F:\_python") for file in f]
    print(x)
    
    >>> ["F:\_python\dict.py", "F:\_python\progr.txt", "F:\_python\readl.py"]
    

    os.listdir() - get only txt files

     arr_txt = [x for x in os.listdir() if x.endswith(".txt")]
     print(arr_txt)
     
     >>> ["work.txt", "3ebooks.txt"]
    

    Using glob to get the full path of the files

    If I should need the absolute path of the files:

    from path import path
    from glob import glob
    x = [path(f).abspath() for f in glob("F:\*.txt")]
    for f in x:
        print(f)
    
    >>> F:acquistionline.txt
    >>> F:acquisti_2018.txt
    >>> F:ootstrap_jquery_ecc.txt
    

    Using os.path.isfile to avoid directories in the list

    import os.path
    listOfFiles = [f for f in os.listdir() if os.path.isfile(f)]
    print(listOfFiles)
    
    >>> ["a simple game.py", "data.txt", "decorator.py"]
    

    Using pathlib from Python 3.4

    import pathlib
    
    flist = []
    for p in pathlib.Path(".").iterdir():
        if p.is_file():
            print(p)
            flist.append(p)
    
     >>> error.PNG
     >>> exemaker.bat
     >>> guiprova.mp3
     >>> setup.py
     >>> speak_gui2.py
     >>> thumb.PNG
    

    With list comprehension:

    flist = [p for p in pathlib.Path(".").iterdir() if p.is_file()]
    

    Alternatively, use pathlib.Path() instead of pathlib.Path(".")

    Use glob method in pathlib.Path()

    import pathlib
    
    py = pathlib.Path().glob("*.py")
    for file in py:
        print(file)
    
    >>> stack_overflow_list.py
    >>> stack_overflow_list_tkinter.py
    

    Get all and only files with os.walk

    import os
    x = [i[2] for i in os.walk(".")]
    y=[]
    for t in x:
        for f in t:
            y.append(f)
    print(y)
    
    >>> ["append_to_list.py", "data.txt", "data1.txt", "data2.txt", "data_180617", "os_walk.py", "READ2.py", "read_data.py", "somma_defaltdic.py", "substitute_words.py", "sum_data.py", "data.txt", "data1.txt", "data_180617"]
    

    Get only files with next and walk in a directory

     import os
     x = next(os.walk("F://python"))[2]
     print(x)
     
     >>> ["calculator.bat","calculator.py"]
    

    Get only directories with next and walk in a directory

     import os
     next(os.walk("F://python"))[1] # for the current dir use (".")
     
     >>> ["python3","others"]
    

    Get all the subdir names with walk

    for r,d,f in os.walk("F:\_python"):
        for dirs in d:
            print(dirs)
    
    >>> .vscode
    >>> pyexcel
    >>> pyschool.py
    >>> subtitles
    >>> _metaprogramming
    >>> .ipynb_checkpoints
    

    os.scandir() from Python 3.5 and greater

    import os
    x = [f.name for f in os.scandir() if f.is_file()]
    print(x)
    
    >>> ["calculator.bat","calculator.py"]
    
    # Another example with scandir (a little variation from docs.python.org)
    # This one is more efficient than os.listdir.
    # In this case, it shows the files only in the current directory
    # where the script is executed.
    
    import os
    with os.scandir() as i:
        for entry in i:
            if entry.is_file():
                print(entry.name)
    
    >>> ebookmaker.py
    >>> error.PNG
    >>> exemaker.bat
    >>> guiprova.mp3
    >>> setup.py
    >>> speakgui4.py
    >>> speak_gui2.py
    >>> speak_gui3.py
    >>> thumb.PNG
    

    Examples:

    Ex. 1: How many files are there in the subdirectories?

    In this example, we look for the number of files that are included in all the directory and its subdirectories.

    import os
    
    def count(dir, counter=0):
        "returns number of files in dir and subdirs"
        for pack in os.walk(dir):
            for f in pack[2]:
                counter += 1
        return dir + " : " + str(counter) + "files"
    
    print(count("F:\python"))
    
    >>> "F:\python" : 12057 files"
    

    Ex.2: How to copy all files from a directory to another?

    A script to make order in your computer finding all files of a type (default: pptx) and copying them in a new folder.

    import os
    import shutil
    from path import path
    
    destination = "F:\file_copied"
    # os.makedirs(destination)
    
    def copyfile(dir, filetype="pptx", counter=0):
        "Searches for pptx (or other - pptx is the default) files and copies them"
        for pack in os.walk(dir):
            for f in pack[2]:
                if f.endswith(filetype):
                    fullpath = pack[0] + "\" + f
                    print(fullpath)
                    shutil.copy(fullpath, destination)
                    counter += 1
        if counter > 0:
            print("-" * 30)
            print("	==> Found in: `" + dir + "` : " + str(counter) + " files
    ")
    
    for dir in os.listdir():
        "searches for folders that starts with `_`"
        if dir[0] == "_":
            # copyfile(dir, filetype="pdf")
            copyfile(dir, filetype="txt")
    
    
    >>> _compiti18Compito Contabilità 1conti.txt
    >>> _compiti18Compito Contabilità 1modula4.txt
    >>> _compiti18Compito Contabilità 1moduloa4.txt
    >>> ------------------------
    >>> ==> Found in: `_compiti18` : 3 files
    

    Ex. 3: How to get all the files in a txt file

    In case you want to create a txt file with all the file names:

    import os
    mylist = ""
    with open("filelist.txt", "w", encoding="utf-8") as file:
        for eachfile in os.listdir():
            mylist += eachfile + "
    "
        file.write(mylist)
    

    Example: txt with all the files of an hard drive

    """
    We are going to save a txt file with all the files in your directory.
    We will use the function walk()
    """
    
    import os
    
    # see all the methods of os
    # print(*dir(os), sep=", ")
    listafile = []
    percorso = []
    with open("lista_file.txt", "w", encoding="utf-8") as testo:
        for root, dirs, files in os.walk("D:\"):
            for file in files:
                listafile.append(file)
                percorso.append(root + "\" + file)
                testo.write(file + "
    ")
    listafile.sort()
    print("N. of files", len(listafile))
    with open("lista_file_ordinata.txt", "w", encoding="utf-8") as testo_ordinato:
        for file in listafile:
            testo_ordinato.write(file + "
    ")
    
    with open("percorso.txt", "w", encoding="utf-8") as file_percorso:
        for file in percorso:
            file_percorso.write(file + "
    ")
    
    os.system("lista_file.txt")
    os.system("lista_file_ordinata.txt")
    os.system("percorso.txt")
    

    All the file of C: in one text file

    This is a shorter version of the previous code. Change the folder where to start finding the files if you need to start from another position. This code generate a 50 mb on text file on my computer with something less then 500.000 lines with files with the complete path.

    import os
    
    with open("file.txt", "w", encoding="utf-8") as filewrite:
        for r, d, f in os.walk("C:\"):
            for file in f:
                filewrite.write(f"{r + file}
    ")
    

    How to write a file with all paths in a folder of a type

    With this function you can create a txt file that will have the name of a type of file that you look for (ex. pngfile.txt) with all the full path of all the files of that type. It can be useful sometimes, I think.

    import os
    
    def searchfiles(extension=".ttf", folder="H:\"):
        "Create a txt file with all the file of a type"
        with open(extension[1:] + "file.txt", "w", encoding="utf-8") as filewrite:
            for r, d, f in os.walk(folder):
                for file in f:
                    if file.endswith(extension):
                        filewrite.write(f"{r + file}
    ")
    
    # looking for png file (fonts) in the hard disk H:
    searchfiles(".png", "H:\")
    
    >>> H:4bs_18Dolphins5.png
    >>> H:4bs_18Dolphins6.png
    >>> H:4bs_18Dolphins7.png
    >>> H:5_18marketing htmlassetsimageslogo2.png
    >>> H:7z001.png
    >>> H:7z002.png
    

    (New) Find all files and open them with tkinter GUI

    I just wanted to add in this 2019 a little app to search for all files in a dir and be able to open them by doubleclicking on the name of the file in the list. enter image description here

    import tkinter as tk
    import os
    
    def searchfiles(extension=".txt", folder="H:\"):
        "insert all files in the listbox"
        for r, d, f in os.walk(folder):
            for file in f:
                if file.endswith(extension):
                    lb.insert(0, r + "\" + file)
    
    def open_file():
        os.startfile(lb.get(lb.curselection()[0]))
    
    root = tk.Tk()
    root.geometry("400x400")
    bt = tk.Button(root, text="Search", command=lambda:searchfiles(".png", "H:\"))
    bt.pack()
    lb = tk.Listbox(root)
    lb.pack(fill="both", expand=1)
    lb.bind("<Double-Button>", lambda x: open_file())
    root.mainloop()
    

    PHP dirname () function dirname: Questions

    os.path.dirname(__file__) returns empty

    2 answers

    I want to get the path of the current directory under which a .py file is executed.

    For example a simple file D: est.py with code:

    import os
    
    print os.getcwd()
    print os.path.basename(__file__)
    print os.path.abspath(__file__)
    print os.path.dirname(__file__)
    

    It is weird that the output is:

    D:
    test.py
    D:	est.py
    EMPTY
    

    I am expecting the same results from the getcwd() and path.dirname().

    Given os.path.abspath = os.path.dirname + os.path.basename, why

    os.path.dirname(__file__)
    

    returns empty?

    174

    Answer #1

    Because os.path.abspath = os.path.dirname + os.path.basename does not hold. we rather have

    os.path.dirname(filename) + os.path.basename(filename) == filename
    

    Both dirname() and basename() only split the passed filename into components without taking into account the current directory. If you want to also consider the current directory, you have to do so explicitly.

    To get the dirname of the absolute path, use

    os.path.dirname(os.path.abspath(__file__))
    

    What is the difference between os.path.basename() and os.path.dirname()?

    2 answers

    What is the difference between os.path.basename() and os.path.dirname()?

    I already searched for answers and read some links, but didn"t understand. Can anyone give a simple explanation?

    158

    Answer #1

    Both functions use the os.path.split(path) function to split the pathname path into a pair; (head, tail).

    The os.path.dirname(path) function returns the head of the path.

    E.g.: The dirname of "/foo/bar/item" is "/foo/bar".

    The os.path.basename(path) function returns the tail of the path.

    E.g.: The basename of "/foo/bar/item" returns "item"

    From: http://docs.python.org/3/library/os.path.html#os.path.basename

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