1

我有一个名为 ***.zip 的 zip 文件。我使用下面的命令来解压缩它。一旦我解压缩，其中的文件也是“Zip”文件（超过 3 个 zip 文件）。您能否让我知道如何解压缩这些文件。

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

试图做——

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$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
4

3 回答 3

2 
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\"
}
于 2018-05-16T14:34:39.337 回答
2

您可以在步骤块中使用管道实用程序步骤，该步骤应该适用于脚本式和声明式管道样式

steps {
      unzip zipFile: 'file.zip', dir: '<directory>'
}
于 2020-02-10T04:19:46.500 回答
0 
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 流水线脚本下使用它来实现上述要求

于 2018-05-16T14:54:16.963 回答