Bash命令行重命名通配符 [英] Bash command-line to rename wildcard

问题描述

在我的 / opt / myapp dir中，我有一个远程的自动化进程，它将删除< anything> ;版本> .zip ，其中< anything> 可以字面上是任何字母数字文件名，其中< version> 将是一个版本号。所以，这个自动化过程将会提供的例子有：

• fizz-0.1.0.zip code>


• buzz-1.12.35.zip


• foo-1.0.0.zip


• bar-3.0.9.RC.zip

等。通过这个问题范围之外的控制，我保证 在任何给定的时间只有一个ZIP文件存在于 / opt / myapp 。我需要编写一个 Bash shell命令来重命名这些文件并将它们移动到 / opt / staging 。对于重命名，ZIP文件需要删除其版本。因此 / opt / myapp /< anything> - < version> .zip 被重命名并移动到 / opt / staging /< anything> ;的.zip 。使用上面的例子：

• /opt/myapp/fizz-0.1.0.zip => /opt/staging/fizz.zip


• /opt/myapp/buzz-1.12 .35.zip => /opt/staging/buzz.zip


• /opt/myapp/foo-1.0.0.zip => /opt/staging/foo.zip


• /opt/myapp/bar-3.0.9.RC.zip => /opt/staging/bar.zip
  $ b

  目录的移动是显而易见的，但重命名却让我把头发拉出来。我需要以某种方式保存< anything> ，然后在命令中重新访问它。 该命令必须是通用的，并且可以不带任何参数。

  
  
  

  到目前为止，我的最佳尝试（甚至没有接近工作） ：

  
  

  file = *。zip;文件= ?; mv file / opt / staging

  
  

  关于如何做到这一点的任何想法？解决方案

 用于* .zip文件。做
 [[-e $ file]] ||继续＃处理zero-match case没有nullglob 
 mv  - $ file/opt/staging/\"${file%-*}.zip
完成
  

$ {file％ - *} > - 在文件名中。因此，我们将 fizz-0.1.0.zip 更改为 fizz ，然后添加一个前导 / opt / staging / 和尾部 .zip 。

---

为了使这个更通用（使用多个扩展），请看下面的函数（可调用为一个命令;函数体也可以放在脚本中，使用＃！/ bin / bash shebang，如果删除了 local 声明）：

  stage（）{
本地文件ext 
用于文件;做
 [[-e $ file]] ||继续
 [[$ file = *  -  *。*]] || {
 printf'错误：文件名％q不包含短划线和点'\\''$ file>& 2 
 continue 
} 
 ext = $ {file ## *。} 
 mv  - $ file/opt/staging/\"${file%-*}.$ext
 done 
} 
  

...定义了该函数，您可以运行：

  stage * .zip * .txt 
  

...或您选择的任何其他模式。

In my /opt/myapp dir I have a remote, automated process that will be dropping files of the form <anything>-<version>.zip, where <anything> could literally be any alphanumeric filename, and where <version> will be a version number. So, examples of what this automated process will be delivering are:

• fizz-0.1.0.zip
• buzz-1.12.35.zip
• foo-1.0.0.zip
• bar-3.0.9.RC.zip

etc. Through controls outside the scope of this question, I am guaranteed that only one of these ZIP files will exist under /opt/myapp at any given time. I need to write a Bash shell command that will rename these files and move them to /opt/staging. For the rename, the ZIP files need to have their version dropped. And so /opt/myapp/<anything>-<version>.zip is renamed and moved to /opt/staging/<anything>.zip. Using the examples above:

• /opt/myapp/fizz-0.1.0.zip => /opt/staging/fizz.zip
• /opt/myapp/buzz-1.12.35.zip => /opt/staging/buzz.zip
• /opt/myapp/foo-1.0.0.zip => /opt/staging/foo.zip
• /opt/myapp/bar-3.0.9.RC.zip => /opt/staging/bar.zip

The directory move is obvious and easy, but the rename is making me pull my hair out. I need to somehow save off the <anything> and then re-access it later on in the command. The command must be generic and can take no arguments.

My best attempt (which doesn't even come close to working) so far is:

file=*.zip; file=?; mv file /opt/staging

Any ideas on how to do this?

解决方案

for file in *.zip; do
  [[ -e $file ]] || continue # handle zero-match case without nullglob
  mv -- "$file" /opt/staging/"${file%-*}.zip"
done

${file%-*} removes everything after the last - in the filename. Thus, we change fizz-0.1.0.zip to fizz, and then add a leading /opt/staging/ and a trailing .zip.

---

To make this more generic (working with multiple extensions), see the following function (callable as a command; function body could also be put into a script with a #!/bin/bash shebang, if one removed the local declarations):

stage() {
  local file ext
  for file; do
    [[ -e $file ]] || continue
    [[ $file = *-*.* ]] || {
      printf 'ERROR: Filename %q does not contain a dash and a dot\n' "$file" >&2
      continue
    }
    ext=${file##*.}
    mv -- "$file" /opt/staging/"${file%-*}.$ext"
  done
}

...with that function defined, you can run:

stage *.zip *.txt

...or any other pattern you so choose.

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