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Parsing Linux multi-application docker automatic deployment script

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Release: 2024-06-05 13:47:05
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摘要: Linux 多应用程序docker自动部署脚本可以结合jenkins分布式部署

解析Linux 多应用程序docker自动部署脚本

Linux 多应用程序docker自动部署脚本可以结合jenkins分布式部署

参数:

镜像名:端口的格式:版本号

例如:spring-client:8764:0.0.1

1.先用maven install 打包项目到target目录下

2.项目对应的Dockerfile路径为:/src/main/docker/Dockerfile

3.脚本自动复制Dockerfile和程序文件自动打包镜像部署

4.docker命令运行时加入--restart unless-stopped 可以容器随docker自启

project=("spring-client:8764:0.0.1" "spring-system:8770:0.0.1" "spring-eureka:8761:0.0.1")
mainDir=/opt/jenkins
jenkinsDir=/root/.jenkins/workspace/spring-boot-xinjiang
#判断主文件夹是否存在,不存在创建
if [ ! -d $mainDir ]; then
  mkdir $mainDir
fi
#循环数组创建每个项目的文件夹
for var in ${project[@]} 
do
  OLD_IFS="$IFS"  
  IFS=":"  
  array=($var)  
  IFS="$OLD_IFS"
  name=${array[0]}
  port=${array[1]}
  version=${array[2]}  
  echo 项目名称$name;
  if [ ! -d $mainDir/$name ]; then
    echo "项目目录不存在,创建项目目录"
    mkdir $mainDir"/"$name
  fi
  cd $jenkinsDir;
  jarName=$name/target/$name-1.0.0.jar
  if [ -f $jarName ]; then
    echo "jar源文件存在,正在复制"
    #复制程序文件(war/jar)的名称
    cp $jarName $mainDir/$name/$name-$version-SNAPSHOT.jar
  else
    echo "jar源文件不存在"
  fi
  #Dockerfile文件的路径
  dockerFileName=$name/src/main/docker/Dockerfile
  if [ -f $dockerFileName ]; then
    echo "docker文件存在,正在复制"
    #复制Dockerfile程序文件的名称
    cp $dockerFileName $mainDir/$name
  else
    echo "docker文件不存在"
  fi
  if [ -f $dockerFileName ] && [ -f $jarName ]; then
    cd $mainDir/$name
    if docker ps -a|grep -i $name;then
      docker stop $name
      docker rm $name
    fi
    imagesid=`docker images|grep -i $name | awk '{print $3}'`

    if [ "$imagesid" == "" ];then
      echo  "镜像不存在"
    else
      echo  "镜像存在删除后构建"
      if docker ps -a|grep -i $name;then
       echo "镜像容器存在,正在停止容器"
       docker stop $name
       echo "镜像容器删除"
       docker rm $name
      fi
      docker rmi $imagesid -f
    fi
     docker build -t $name .
     echo "镜像构建成功"
     echo "容器构建中"
     docker run -d -p $port:$port --name $name --restart unless-stopped $dockerName $name
     echo "容器启动成功"
  else
    echo "镜像构建失败"
  fi
done
exit 0 
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source:linuxprobe.com
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