Docker shared container within multiple docker compose projects
I have a docker-compose.yml which get's 2 services up (I have left out all irrelevant data from it).
app:
build:
context: .
dockerfile: ./docker/app/Dockerfile
image: ...
container_name: app-${ENV}
depends_on:
- db
expose:
- 80
db:
image: ...
container_name: my-cool-db
ports:
- "3306:3306"
The point to see here is that app is getting a container name depending on the parameter. So basically I have a shell script running the following:
ENV=$1 docker-compose -p $1 up -d
So in short, whatever I forward as parameter, the new app container should be brought up. For example if I do sh initializer.sh first I will get app-first container. -p parameter is specified so I can have multiple instances of same container classified as a different project.
If I have a single container this works great, and I end up with say:
app-first
app-second
app-third
What I would like to achieve is to have all containers use the same DB. But when I do a docker-compose my DB container still wants to be brought up independent of his existence already.
Is it an issue that it tries to create a DB under different project name, but with same container name so it causes the collision?
Can this be made without bringing up 2 separate DB containers?
docker docker-compose dockerfile
edited Nov 16 '18 at 15:31
Siyu
3,20411231
asked Nov 16 '18 at 11:16
NorgulNorgul
1,64011845
add a comment |
I have a docker-compose.yml which get's 2 services up (I have left out all irrelevant data from it).
app:
build:
context: .
dockerfile: ./docker/app/Dockerfile
image: ...
container_name: app-${ENV}
depends_on:
- db
expose:
- 80
db:
image: ...
container_name: my-cool-db
ports:
- "3306:3306"
The point to see here is that app is getting a container name depending on the parameter. So basically I have a shell script running the following:
ENV=$1 docker-compose -p $1 up -d
So in short, whatever I forward as parameter, the new app container should be brought up. For example if I do sh initializer.sh first I will get app-first container. -p parameter is specified so I can have multiple instances of same container classified as a different project.
If I have a single container this works great, and I end up with say:
app-first
app-second
app-third
What I would like to achieve is to have all containers use the same DB. But when I do a docker-compose my DB container still wants to be brought up independent of his existence already.
Is it an issue that it tries to create a DB under different project name, but with same container name so it causes the collision?
Can this be made without bringing up 2 separate DB containers?
docker docker-compose dockerfile
edited Nov 16 '18 at 15:31
Siyu
3,20411231
asked Nov 16 '18 at 11:16
NorgulNorgul
1,64011845
add a comment |
I have a docker-compose.yml which get's 2 services up (I have left out all irrelevant data from it).
app:
build:
context: .
dockerfile: ./docker/app/Dockerfile
image: ...
container_name: app-${ENV}
depends_on:
- db
expose:
- 80
db:
image: ...
container_name: my-cool-db
ports:
- "3306:3306"
The point to see here is that app is getting a container name depending on the parameter. So basically I have a shell script running the following:
ENV=$1 docker-compose -p $1 up -d
So in short, whatever I forward as parameter, the new app container should be brought up. For example if I do sh initializer.sh first I will get app-first container. -p parameter is specified so I can have multiple instances of same container classified as a different project.
If I have a single container this works great, and I end up with say:
app-first
app-second
app-third
What I would like to achieve is to have all containers use the same DB. But when I do a docker-compose my DB container still wants to be brought up independent of his existence already.
Is it an issue that it tries to create a DB under different project name, but with same container name so it causes the collision?
Can this be made without bringing up 2 separate DB containers?
docker docker-compose dockerfile
edited Nov 16 '18 at 15:31
Siyu
3,20411231
asked Nov 16 '18 at 11:16
NorgulNorgul
1,64011845
I have a docker-compose.yml which get's 2 services up (I have left out all irrelevant data from it).
app:
build:
context: .
dockerfile: ./docker/app/Dockerfile
image: ...
container_name: app-${ENV}
depends_on:
- db
expose:
- 80
db:
image: ...
container_name: my-cool-db
ports:
- "3306:3306"
The point to see here is that app is getting a container name depending on the parameter. So basically I have a shell script running the following:
ENV=$1 docker-compose -p $1 up -d
So in short, whatever I forward as parameter, the new app container should be brought up. For example if I do sh initializer.sh first I will get app-first container. -p parameter is specified so I can have multiple instances of same container classified as a different project.
If I have a single container this works great, and I end up with say:
app-first
app-second
app-third
What I would like to achieve is to have all containers use the same DB. But when I do a docker-compose my DB container still wants to be brought up independent of his existence already.
Is it an issue that it tries to create a DB under different project name, but with same container name so it causes the collision?
Can this be made without bringing up 2 separate DB containers?
docker docker-compose dockerfile
docker docker-compose dockerfile
edited Nov 16 '18 at 15:31
Siyu
3,20411231
asked Nov 16 '18 at 11:16
NorgulNorgul
1,64011845
edited Nov 16 '18 at 15:31
Siyu
3,20411231
asked Nov 16 '18 at 11:16
NorgulNorgul
1,64011845
edited Nov 16 '18 at 15:31
Siyu
3,20411231
edited Nov 16 '18 at 15:31
Siyu
3,20411231
edited Nov 16 '18 at 15:31
Siyu
3,20411231
3,20411231
asked Nov 16 '18 at 11:16
NorgulNorgul
1,64011845
asked Nov 16 '18 at 11:16
NorgulNorgul
1,64011845
asked Nov 16 '18 at 11:16
NorgulNorgul
1,64011845
1,64011845
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
A hacky solution:
change your compose file to
services:
app:
image: ...
container_name: app-${ENV}
networks:
- shared
expose:
- 80
db:
image: ...
container_name: my-cool-db
networks:
- shared
ports:
- "3306:3306"
networks:
shared:
external: true
Then first create the network docker network create shared
Bring up db: docker-compose up -d db
First app: ENV=first docker-compose -p first up -d app
Second app: ENV=second docker-compose -p second up -d app
answered Nov 16 '18 at 13:27
SiyuSiyu
3,20411231
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
A hacky solution:
change your compose file to
services:
app:
image: ...
container_name: app-${ENV}
networks:
- shared
expose:
- 80
db:
image: ...
container_name: my-cool-db
networks:
- shared
ports:
- "3306:3306"
networks:
shared:
external: true
Then first create the network docker network create shared
Bring up db: docker-compose up -d db
First app: ENV=first docker-compose -p first up -d app
Second app: ENV=second docker-compose -p second up -d app
answered Nov 16 '18 at 13:27
SiyuSiyu
3,20411231
add a comment |
A hacky solution:
change your compose file to
services:
app:
image: ...
container_name: app-${ENV}
networks:
- shared
expose:
- 80
db:
image: ...
container_name: my-cool-db
networks:
- shared
ports:
- "3306:3306"
networks:
shared:
external: true
Then first create the network docker network create shared
Bring up db: docker-compose up -d db
First app: ENV=first docker-compose -p first up -d app
Second app: ENV=second docker-compose -p second up -d app
answered Nov 16 '18 at 13:27
SiyuSiyu
3,20411231
add a comment |
A hacky solution:
change your compose file to
services:
app:
image: ...
container_name: app-${ENV}
networks:
- shared
expose:
- 80
db:
image: ...
container_name: my-cool-db
networks:
- shared
ports:
- "3306:3306"
networks:
shared:
external: true
Then first create the network docker network create shared
Bring up db: docker-compose up -d db
First app: ENV=first docker-compose -p first up -d app
Second app: ENV=second docker-compose -p second up -d app
answered Nov 16 '18 at 13:27
SiyuSiyu
3,20411231
A hacky solution:
change your compose file to
services:
app:
image: ...
container_name: app-${ENV}
networks:
- shared
expose:
- 80
db:
image: ...
container_name: my-cool-db
networks:
- shared
ports:
- "3306:3306"
networks:
shared:
external: true
Then first create the network docker network create shared
Bring up db: docker-compose up -d db
First app: ENV=first docker-compose -p first up -d app
Second app: ENV=second docker-compose -p second up -d app
answered Nov 16 '18 at 13:27
SiyuSiyu
3,20411231
edited Nov 16 '18 at 13:36
answered Nov 16 '18 at 13:27
SiyuSiyu
3,20411231
answered Nov 16 '18 at 13:27
SiyuSiyu
3,20411231
answered Nov 16 '18 at 13:27
SiyuSiyu
3,20411231
3,20411231
add a comment |
add a comment |
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