I created an image called "helloworld" based on some project I have.
If I run :
docker images
I can see it there on top of the list.
Now if I want to run it, docker complains it does not exist.
Running this :
docker run -p 8080:8080 helloworld
returns this :
docker: Error response from daemon: pull access denied for helloworld,
repository does not exist or may require 'docker login': denied:
requested access to the resource is denied. See 'docker run --help'.
Why is docker complaining that my image does not exist?
It complains, because this:
docker run -p 8080:8080 helloworld
is a shortened version of this:
docker run -p 8080:8080 docker.io/library/helloworld:latest
and, based on the screenshot, your image is called
docker.io/library/helloworld:1.0
so the proper command (skipping default prefix) is:
docker run -p 8080:8080 helloworld:1.0
Other answers are correct but don't explain clearly why they are correct.
The docker run command expects an image name and, optionally, a tag (i.e., docker run IMAGE[:TAG]). If a tag is not provided, the default is latest.
Running docker run helloworld is equivalent to docker run helloworld:latest, but this image does not exist in the OP's system according to docker images.
The solution is to specify the image tag: helloworld:1.0.
Your image build command:
docker build --rm -f "Dockerfile" -t helloworld:1.0 "."
Meaning your image name is helloworld:1.0
If you had run docker build --rm -f "Dockerfile" -t helloworld "."
Then your image name is helloworld or helloworld:latest and your first try would have worked.
To solve your issue run this instead:
docker run -p 8080:8080 helloworld:1.0
Related
i have creted a docker volume with such command
docker run -ti --rm -v TestVolume1:/testvolume1 ubuntu
then i created a file there, called TestFile.txt and added text to it
Also i have a simple "Hello world" .net core app with Dockerfile
FROM mcr.microsoft.com/dotnet/aspnet:6.0
COPY bin/Release/net6.0/publish/ ShareFileTestInstance1/
WORKDIR /ShareFileTestInstance1
ENTRYPOINT ["dotnet", "ShareFileTestInstance1.dll"]
I published it using
dotnet publish -c Release
then ran
docker build -t counter-image -f Dockerfile .
And finally executed
docker run -it --rm --name=counter-container counter-image -v TestVolume1:/testvolume1 ubuntu
to run my app with a docker volume
So what i want to achive to access a file which is in a volume("TestFile.txt" in my case) from a code in the container.
for example
Console.WriteLine(File.Exists("WHAT FILE PATH HAS TO BE HERE") ? "File exists." : "File does not exist.");
Is it also possible to combine all this stuff in a Dockerfile? I want to add one more container next and connect to the volume to save data there.
The parameters for docker run can be either for docker or for the program running in the docker container. Parameters for docker go before the image name and parameters for the program in the container go after the image name.
The volume mapping is a parameter for docker, so it should go before the image name. So instead of
docker run -it --rm --name=counter-container counter-image -v TestVolume1:/testvolume1 ubuntu
you should do
docker run -it --rm --name=counter-container -v TestVolume1:/testvolume1 counter-image
When you do that, your file should be accessible for your program at /testvolume1/TestFile.txt.
It's not possible to do the mapping in the Dockerfile as you ask. Mappings may vary from docker host to docker host, so they need to be specified at run-time.
I was following this guide on customizing MySQL databases in Docker, and ran this command multiple times after making tweaks to the mounted sql files:
docker run -d -p 3306:3306 --name my-mysql -v /Users/pneedham/dev/docker-testing/sql-scripts:/docker-entrypoint-initdb.d/ -e MYSQL_ROOT_PASSWORD=supersecret -e MYSQL_DATABASE=company mysql
On all subsequent executions of that command, I would see an error like this:
docker: Error response from daemon: Conflict. The container name "/my-mysql" is already in use by container "9dc103de93b7ad0166bb359645c12d49e0aa4a3f2330b5980e455cec24843663". You have to remove (or rename) that container to be able to reuse that name.
See 'docker run --help'.
What I'd like to know is whether that docker run command can be modified to auto-kill the previous container (if it exists)? Or if there is a different command that has the same desired result.
If I were to create a shell script to do that for me, I'd first run docker ps -aqf "name=mysql" and if there is any output, use that resulting container ID by running docker rm -f $containerID. And then run the original command.
docker run command has a --rm arguments that deletes the container after the run is completed. see the docs . So, just change your command to
docker run --rm -d -p 3306:3306 --name my-mysql -v /Users/pneedham/dev/docker-testing/sql-scripts:/docker-entrypoint-initdb.d/ -e MYSQL_ROOT_PASSWORD=supersecret -e MYSQL_DATABASE=company mysql
i would like to run the docker image gcr.io/google-containers/echoserver:1.8. i tried this command docker run -d gcr.io/google-containers/echoserver:1.8 -p 8081:80 but it seems not to work. there is the error message:
and when i insert the docker ps command, no process is running. i don't know if parameters are placed badly or it is not the right command to run this image
All the strings after the image name is considered as a command to run in the container:
you may use:
docker run -d -p 8081:80 gcr.io/google-containers/echoserver:1.8
then the container will run the Entrypoint/CMD wich is configured in the image
I'm trying to create a new docker container using the following command:
docker run -d -it --name compsci -v /c/Users/garre/Documents/CPSC_Courses:/homechapmanfse/computing-resources:cs_base
However, it gives me this error message:
"docker run" requires at least 1 argument.
See 'docker run --help'.
Usage: docker run [OPTIONS] IMAGE [COMMAND] [ARG...]
Run a command in a new container
How would I fix this?
You have to provide the name of the image that you want to run. This is currently missing in your command.
For example, if I were to run mysql, I would execute this:
docker run --name some-mysql -e MYSQL_ROOT_PASSWORD=my-secret-pw -d mysql
See the last argument, mysql? That is the name of the image.
Think that you it has build image in your machine. You must inform name of image run.
docker run image-name
This command --name is necessary only you specific name for your container. And the -it command must be entered only when entering the executed container.
docker run -d -it -v
/c/Users/garre/Documents/CPSC_Courses:/homechapmanfse/computing-resources:cs_base
--name 'the name you want to give' 'official name of the image'
Description
I got error "Error response from daemon: invalid mount config for type "volume": invalid mount path" in Windows Container
Steps to reproduce the issue:
1. DockerFile
FROM microsoft/aspnetcore-build AS base
WORKDIR /app
ENTRYPOINT [ "dotnet", "Test.dll" ]
Run command docker build -t docker-vol .
Run Command docker run -it -p 8001:80 -v D:\Projects\Docker\publish:c:/app --name docker-vol-test docker-vol (This works)
Run Command docker run -it -p 8001:80 --mount "source=D:\Projects\Docker\publish,target=c:/app" --name docker-vol-test docker-vol This fails with Error response from daemon: invalid mount config for type "volume": invalid volume name
I am wondering how to work with --mount and whether it is possible to use relative path instead of absolute path with -v?
You are using a bind mount, but because you have not specified a type, then it has defaulted to volume. In this case, source must be the name of the volume, or omitted for an anonymous volume.
Because you have give a path instead, you are getting this error. If you add a type key to your command, it should work:
docker run -it -p 8001:80 --mount 'type=bind, source="D:\Projects\Docker\publish", target="c:/app"' --name docker-vol-test docker-vol
In answer to your second point, bind mounts require an absolute path. The usual way to use a relative path in Linux-land is to prepend the path with $PWD. In Windows, the equivalent of SPWD would be %cd%, so if you were running from D:\Projects\Docker, then the above would probably be:
docker run -it -p 8001:80 --mount 'type=bind, source="%cwd%\publish", target="c:/app"' --name docker-vol-test docker-vol
Note that I have no experience of Docker under Windows, but I believe the above should work.
The above is correct but if you want to use -v instead the syntax is: docker run -v C:\SomePath:C:\app\somePath image
Note the path must exist or the command fails.