fopen does not create files when used with docker

Question

Code of Dockerfile is here

FROM php:7.0-cli
COPY ./ /

Code of create.php is here

<?php
$fp = fopen($argv[1],'wa+');
fwrite($fp,'sample');
fclose($fp);
?>

I build my docker using following command in debian - stretch

docker build -t create-tool .

I run it using following command in debian - stretch

docker run create-tool php create.php /home/user1/Desktop/test

But it is not creating the file, instead it throws the error "failed to open stream".

You are trying to pass in a file location on your host OS. Docker containers can only access file paths within the running container (unless you bind mount them) — Jacob Lambert, Sep 03 '17 at 07:10
Mounting in a Docker file - https://stackoverflow.com/questions/23439126/how-to-mount-host-directory-in-docker-container — Nigel Ren, Sep 03 '17 at 07:53
@Neodan What permissions are required and where can they be given? — naman1994, Jan 28 '21 at 16:17

score 0 · Answer 1 · edited Feb 24 '21 at 05:52

0

How about using touch()? It can solve your problem!

edited Feb 24 '21 at 05:52

mik13ST

answered Sep 03 '17 at 07:17

Mr.Apr

score 0 · Answer 2 · edited Jun 20 '20 at 09:12

3 Scenarios I would investigate :

Because /home/user1/Desktop/test doesn't exist within the docker container. That file exists on your host system, so it's out of reach for the container.

If you want to access files on your host, you'll need to share the folders you want the container to access with a volume.
The path exists in your container, but is for example a directory. You can check quickly if a file exists and/or is accessible within your container by running docker run create-tool ls -lsah /home/user1/Desktop/test.

This will then list what the container can access with the given path.
Double-check what the value of $argv[1] in your PHP script is. You wouldn't want to get escaped characters, etc. print_r($argv); should help.

2 Answers2