shell - 如何在Dockerfile中将环境变量作为Shell脚本的参数传递？

Question

如何在运行时将环境变量作为Dockerfile中的Shell脚本的参数传递？

PUBLISH_HOST来自.env文件

我的docker-compose文件如下所示:

version: '2'
services:
tomcat:
    build:
      context: .
      args:
        PUBLISH_HOST: ${PUBLISH_HOST}
        dockerfile: Dockerfile.tomcat
    container_name: tomcat
    image: tomcat-new:latest
    restart: always
    ports:
      - 8080:8080

Dockerfile如下:

FROM 99taxis/tomcat7
ARG PUBLISH_HOST //taken from docker-compose  for all other this is working fine not working while passing in a shell script 
ENV PUBLISH_HOST $MAXPAAS_PUBLISH_HOST
COPY update-tomcat-path-config.sh /opt/tomcat/webapps
RUN cd /opt/tomcat/webapps && chmod 755 update-tomcat-path-config.sh && ./update-tomcat-path-config.sh ${PUBLISH_HOST}

但是输入的 ${PUBLISH_HOST}在此shell脚本中未被接受为参数。

有什么办法可以解决这个问题？

这是我在尝试 docker-compose up时收到的错误消息

Step 25/37 : /opt/tomcat/webapps && chmod 755 update-tomcat-path-config.sh && /update-tomcat-path-config.sh ${PUBLISH_HOST} 
 ---> Running in 8ad3b1ecd73e

###################
Script starts Here:
###################

No parameter to the script is given, please enter it....!!!!

如何解决呢？

Answer 1

参见“Setting Default Docker Environment Variables During Image Build”

您的行应为:

ARG PUBLISH_HOST //taken from docker-compose for all other this is working fine not working while passing in a shell script
ENV PUBLISH_HOST $PUBLISH_HOST 

这将允许 ENV在构建时被 ARG值覆盖:

docker build --build-arg PUBLISH_HOST=aValue

对于 docker-compose build

docker-compose build --build-arg PUBLISH_HOST=aValue aSERVICE

注意:

您的 docker-compose.yml可能不使用您的dockerfile，因为 dockerfile: Dockerfile.tomcat没有正确缩进。
参见 docker-compose.yml "alternate Dockerfile":

Compose uses an alternate file to build with.
A build path must also be specified.

build:
  context: .
  dockerfile: Dockerfile-alternate