How to open a random Image from a given file directory?

Question

I am trying to open the image from this directory but am not been able to. It gives me the following error:

WindowsError: [Error 123] The filename, directory name, or volume label syntax is incorrect.

This is my code:

import os, random
random.choice(os.listdir("C:\\Users\rkp10\AppData\Local\Google\Chrome\User Data\Default\Extensions\laookkfknpbbblfpciffpaejjkokdgca\0.91.6_0\backgrounds"+'png'))

In here , in the folder backgrounds there are many images.I want a random Image to open.But while running the program , I am getting WindowsError.

What am I doing wrong?

Edit 1

I tried this:

random.choice(os.listdir(r"C:\Users\rkp10\AppData\Local\Google\Chrome\User Data\Default\Extensions\laookkfknpbbblfpciffpaejjkokdgca\0.91.6_0\backgrounds"+'png'))

But am getting the error:

WindowsError: [Error 123] The filename, directory name, or volume label syntax is incorrect: 'C:\Users\rkp10\AppData\Local\Google\Chrome\User Data\Default\Extensions\laookkfknpbbblfpciffpaejjkokdgca\0.91.6_0\backgrounds\*.png/.'

Edit 2

I tried this:

import os, random

a=random.choice(os.listdir(r"C:\Users\rkp10\AppData\Local\Google\Chrome\User Data\Default\Extensions\laookkfknpbbblfpciffpaejjkokdgca\0.91.6_0\backgrounds"))
os.open(a)

Now I dont get error but it does not open the image either.

Edit 3

I have also tried:

 import random,os
folder= "C:\\Users\rkp10\AppData\Local\Google\Chrome\User Data\Default\Extensions\laookkfknpbbblfpciffpaejjkokdgca\0.91.6_0\backgrounds"
a=random.choice(os.listdir(folder))
print(a)

from PIL import Image
file = folder+'\\'+a
Image.open(file).show()

    #os.open(a, os.O_RDWR)
    from PIL import Image
    file = folder+'\\'+a
    Image.open(file).show()

But got the following error once again:

Traceback (most recent call last): File "G:\Grade 12 Project\auto.py", line 4, in a=random.choice(os.listdir(folder)) WindowsError: [Error 123] The filename, directory name, or volume label syntax is incorrect: 'C:\Users\rkp10\AppData\Local\Google\Chrome\User Data\Default\Extensions\laookkfknpbbblfpciffpaejjkokdgca'

Here(the yellow highlighted part) is the directory where my images are stored.

You have unescaped back slashes in there. Unless Python handles that differently, you'll need to add a second backslash after each existing backslash, or make them forward slashes. — Carcigenicate, Dec 06 '16 at 15:49
@Carcigenicate, I gave a try on what you have said but got the error (I have edited) — White Shadow, Dec 07 '16 at 07:17
The edit 2 works fine for me, are you sure that the directory structure you are given doesn't contain any typos ? — ABcDexter, Dec 07 '16 at 09:06
No, it didn't. But something else worked, writing that as an answer. — ABcDexter, Dec 07 '16 at 09:52

score 4 · Accepted Answer · edited May 23 '17 at 10:30

4

Use the PIL instead.

import os, random

folder=r"D:\Study\SO"

a=random.choice(os.listdir(folder))
print(a)

#os.open(a, os.O_RDWR)
from PIL import Image
file = folder+'\\'+a
Image.open(file).show()

source: Open and display .png file in python using PIL

The problem with this is that a doesn't have absolute path to that chosen random files.

In Edit 1, the "png" gets concatenated but there is no folder named "backgroundspng"
In Edit 2, you haven't given the flags for os.open() which can be found here.
In Edit 3, please make sure that you are using the r before string.

In you case, use this :

folder = r"C:\\Users\rkp10\AppData\Local\Google\Chrome\User Data\Default\Extensions\laookkfknpbbblfpciffpaejjkokdgca\0.91.6_0\backgrounds"

edited May 23 '17 at 10:30

Community

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1

answered Dec 07 '16 at 10:08

ABcDexter

2,751
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I have tried and gave my result up in the edit , please check it.Thanks for your time. – White Shadow Dec 07 '16 at 11:46
@WhiteShadow wait, you have a folder before backgrounds, right ? – ABcDexter Dec 07 '16 at 11:48
You are not using right path to the folder, that is it. – ABcDexter Dec 07 '16 at 11:51
Yes, I do have a folder. – White Shadow Dec 07 '16 at 11:51
But in the edit, the path you have written is incorrect. – ABcDexter Dec 07 '16 at 11:52
`C:\\Users\rkp10\AppData\Local\Google\Chrome\User Data\Default\Extensions\laookkfknpbbblfpciffpaejjkokdgca\backgrounds` it should be `...ca\0.91.6_0\backgrounds` There is a folder named "0.91.6_0" – ABcDexter Dec 07 '16 at 11:52
Yeah mate I did did that again and edited my answer too.But still the problem persists. Come on Indian homeboy , help me out xD. ;) – White Shadow Dec 07 '16 at 11:57
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/130010/discussion-between-white-shadow-and-abcdexter). – White Shadow Dec 07 '16 at 12:01

score 0 · Answer 2 · answered Dec 06 '16 at 15:56

0

You need to use a raw string by prepending the string with r:

random.choice(os.listdir(r"C:\Users\rkp10\AppData\Local\Google\Chrome\User Data\Default\Extensions\laookkfknpbbblfpciffpaejjkokdgca\0.91.6_0\backgrounds"+'png'))

Else, you would need to escape each backslash by another backslash.

answered Dec 06 '16 at 15:56

Pierre Barre

2,174
1
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Hey! I am getting this error now `WindowsError: [Error 123] The filename, directory name, or volume label syntax is incorrect: 'C:\\Users\\rkp10\\AppData\\Local\\Google\\Chrome\\User Data\\Default\\Extensions\\laookkfknpbbblfpciffpaejjkokdgca\\0.91.6_0\\backgrounds\\*.png/*.*` I dont know what's wrong? I just want it to open any random image from the directory I gave. – White Shadow Dec 07 '16 at 07:13

How to open a random Image from a given file directory?

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2 Answers2