使用 Jenkins 管道解压缩文件

Question

我有一个名为 ***.zip 的 zip 文件。我使用下面的命令来解压缩它。一旦我解压，里面的文件也是"Zip"个文件（超过3个zip文件）。你能告诉我如何解压缩这些文件吗？

unzip zipFile: "$project_version",dir:"D:\jenkins\DEV\extract\project", quiet: true

正在努力-

unzip dir: 'D:\jenkins\DEV\extract\project', glob: '', zipFile: 'D:\jenkins\DEV\extract\project\project_*.zip'

错误日志

java.io.IOException: D:\jenkins\DEV\extract\project\project_*.zip does not exist.
    at org.jenkinsci.plugins.pipeline.utility.steps.zip.UnZipStepExecution.run(UnZipStepExecution.java:77)
    at org.jenkinsci.plugins.workflow.steps.AbstractSynchronousNonBlockingStepExecution.call(AbstractSynchronousNonBlockingStepExecution.java:47)
    at hudson.security.ACL.impersonate(ACL.java:260)
    at org.jenkinsci.plugins.workflow.steps.AbstractSynchronousNonBlockingStepExecution.run(AbstractSynchronousNonBlockingStepExecution.java:44)
    at java.util.concurrent.Executors$RunnableAdapter.call(Executors.java:511)
    at java.util.concurrent.FutureTask.run(FutureTask.java:266)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1142)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617)
    at java.lang.Thread.run(Thread.java:745)
Finished: FAILURE

解压主 zip 文件后我的目录下的文件。

05/16/2018  04:31 PM    <DIR>          .
05/16/2018  04:31 PM    <DIR>          ..
05/15/2018  12:51 PM           265,637 project-project1_1.0.0.24_bdd86e0c.zip
05/15/2018  12:51 PM         7,924,188 project-project2_1.4.0.130_43dce5e4.zip
05/15/2018  12:51 PM         6,862,842 project-project3_1.0.0.207_c7d5d471.zip
               3 File(s)     15,052,667 bytes
               2 Dir(s)  432,451,330,048 bytes free

我在 windows-

中需要类似的命令

for file in `ls 123_*.zip'; do unzip $file -d `echo $file | cut -d "." -f 1`; done

Answer 1

Add-Type -AssemblyName System.IO.Compression.FileSystem

function Unzip
{
    param([string]$zipfile, [string]$outpath)
    [System.IO.Compression.ZipFile]::ExtractToDirectory($zipfile, $outpath)
}

Unzip "C:\Temp\Powershell\Test.zip" "C:\Temp\Powershell\"
$DirName =  Get-ChildItem "C:\Temp\Powershell\Test"

foreach ( $item in $DirName){
    Unzip "C:\Temp\Powershell\Test$item" "C:\Temp\Powershell\Test\"
}

Answer 2

bat 'for /R . %I in ("*.zip") do ( "C:\Program Files\7-Zip\7z.exe" x -y -o"%~dpnI" "%~fI" )'

我们可以在 Jenkins 管道脚本下使用它来实现上述要求

Answer 3

您可以在步骤块中使用管道实用程序步骤应该同时使用脚本化和声明式管道

steps {
      unzip zipFile: 'file.zip', dir: '<directory>'
}

使用 Jenkins 管道解压缩文件

unzip a file using Jenkins pipeline

unzip

jenkins-groovy

jenkins-pipeline