Question

我有两个bash脚本。从第一个调用第二个。这是第二个，名为build_with_docker.sh：

#!/bin/bash

VERSION=$1
BUILD_ID=$2

echo "-- build/platform/build_with_docker.sh --"
echo VERSION:$VERSION
echo BUILD_ID:$BUILD_ID
# override params
if [ -z "$VERSION" ]
then
    echo "Failed to pass version :("
    exit 1
fi

if [ -z "$BUILD_ID" ]
then
    echo "Failed to pass BUILD_ID :("
fi

SOURCEID=$(docker build -q -f - --build-arg semver=$VERSION --build-arg build_id=$BUILD_ID $GOPATH/src/myproject < $DEVOPS/build/platform/Dockerfile.build)
echo $SOURCEID

我想从第一个脚本中调用上述脚本。这是第一个：

#!/bin/bash

VERSION=${1:-${VERSION}}
if [ -z "$VERSION" ]
then
    echo "VERSION not defined in shell session, nor passed in as parameter"
    exit 1
fi

if [ -z "$BUILD_ID" ]
then
    BUILD_ID=$(date +%s)
    echo "Setting build id to current date"
fi

# Builds, and outputs image to Docker's local image store
# example "sha256:59d19c43c86865aef8..."
DOCKER_IMAGE_ID=`$DEVOPS/build/platform/build_with_docker.sh ${VERSION} ${BUILD_ID}`
echo $DOCKER_IMAGE_ID

当我执行第一个脚本时，第二个脚本将永远不会执行，并且$DOCKER_IMAGE_ID为空。如果我删除传递给第二个脚本的两个参数${VERSION}和${BUILD_ID}，则它可以工作。

编辑：$ DEVOPS是已经设置的环境变量

请告知。

谢谢

在另一个bash脚本中运行bash脚本并传递变量

0 个答案: