使用bash脚本文件重命名 [英] Rename files using bash script

问题描述

我有一个包含大量照片的文件夹中执行bash脚本设置，依次缝合他们到基于用户输入的全景。一旦生成完成，输出文件重命名为 $ X-pano.jpg 和移动一个文件夹更高。

I have a bash script set that executes in a folder containing a number of photos and sequentially stitches them into panoramas based on the user input. Once the generation is completed, the output file is renamed to $x-pano.jpg and moved one folder higher.

我的问题是多少preFIX是基于脚本的顺序执行，这意味着所有文件获得改名 1 pano.jpg 到<$ C $基于脚本执行过程中产生的全景图的数量C>正pano.jpg 。

My issue is the number prefix is based on the sequential execution of the script, meaning all files get renamed 1-pano.jpg to n-pano.jpg based on the number of panoramas generated during the script execution.

我如何修改重命名过程看存储文件夹，并获得最大的 $ X ？我想通过1递增，这个数字和文件的数值preFIX使用。我现在的code是

How can I modify the renaming process to look at the storage folder and get the largest $x? I want to increment that number by 1 and use as the file's numerical prefix. My current code is

//get the list of files in directory and sort in increasing order
$filelist=$(find ../ -maxdepth 1 -type f | sort -n)

//get number of files
$length=${filelist[@]}

//get the last file
$lastFile=${fileList[$((length-1))]}

将得到的文件的列表，排序在递增的顺序，并从列表中的最后一个文件。这就是我会被卡住。使用 - 作为分隔符，我怎么能捕获当前值

will get a list of the files, sort in increasing order and get the last file from the list. This is where I get stuck. Using - as a delimiter, how can I capture the current value?

推荐答案

您可能正在寻找的 $ {VAR %%富} 结构，这条最长的富从变量的末尾

You're probably looking for the ${var%%foo} construct, which strips the longest foo from the end of a variable:

$ F=1-pano.jpg
$ echo ${F%%-pano.jpg}
1
$ F=1222-pano.jpg
$ echo ${F%%-pano.jpg}
1222

尝试 lastNum = $ {lastFile %% - pano.jpg}

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