Question

我有一个名为***。zip的zip文件。我使用下面的命令解压缩它。解压缩后，其中的文件也是“Zip”文件（超过3个zip文件）。能不能让我知道如何解压这些文件。

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

尝试做 -

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

错误记录

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$1$1.call(AbstractSynchronousNonBlockingStepExecution.java:47)
    at hudson.security.ACL.impersonate(ACL.java:260)
    at org.jenkinsci.plugins.workflow.steps.AbstractSynchronousNonBlockingStepExecution$1.run(AbstractSynchronousNonBlockingStepExecution.java:44)
    at java.util.concurrent.Executors$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

解压缩主zip文件后，

我目录下的文件。

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

我需要在Windows中使用类似的命令 -

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

您可以在步骤块中使用管道实用程序步骤应当同时使用脚本式和声明式的管道

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

Answer 3

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达到上述要求

使用Jenkins管道解压缩文件

3 个答案: