我有一个Jenkinsfile，我需要通过连接变量和字符串来准备zip文件路径。我尝试了下面的片段，但它分裂成多行，像这样：

13:14:02  Zip File Path: /apps/external/5/jenkins-node-home/workspace/APPL004316/APIN004621
13:14:02  /my_project
13:14:02  -0.1.29
13:14:02  -SNAPSHOT.zip

预期输出为：/apps/external/5/jenkins-node-home/workspace/APPL004316/APIN004621/my_project-0.1.29-SNAPSHOT.zip嗅探器：

stage('Zip') {
            steps {
                script {
                    env.VERSION = sh script: 'grep -m 1 version_number ${WORKSPACE}/ComponentInfo.toml | tr -s \' \' | tr -d \'"\' | tr -d "\'" | cut -d\' \' -f3', returnStdout: true
                    echo "Version: ${env.VERSION}"
                    env.PROJECT = sh script: 'echo "${WORKSPACE}"|rev|cut -d\'/\' -f1|cut -d\'@\' -f2|rev', returnStdout: true
                    echo "Project: ${env.PROJECT}"
                    env.BASEPATH = sh script: 'echo ${WORKSPACE%/*}' , returnStdout: true
                    env.ZIP_FILEPATH = "${env.BASEPATH}/${env.PROJECT}-${env.VERSION}-SNAPSHOT.zip"
                    echo "Zip File Path: ${env.ZIP_FILEPATH}"
                }
            }
        }

我也试过使用**"+"**，但得到的结果是一样的：

env.ZIP_FILEPATH = env.BASEPATH+"/"+env.PROJECT+"-"+env.VERSION+"-SNAPSHOT.zip"