bash script - 在目录上迭代合并文件 [英] bash script - iterative merge files on a directory

问题描述

我想要在input_directory中迭代合并文件，并将合并的文件放在output_directory中。假设在input_directory中有file1.txt，file2.txt， file3.txt

我想要输出目录包含以下文件：

• merge1.txt：与file1.txt相同


• merge2.txt：merge file1.txt file2.txt


• merge3.txt：merge file1 .txt file2.txt file3.txt

我几乎没有使用过bash脚本的经验，很明显，迭代器在for循环中，但我不知道如何。

提前谢谢。

解决方案

它附加到输出并在下一次迭代期间使用：

 ＃！/ bin / bash 
 
 touch prev #a temp变量保存以前的文件名
i = 1; 
 
用于$（ls / INPUT / file *）中的文件; 
 do 
 cat / OUTPUT / $ prev / INPUT / $ file> / OUTPUT / merge $ i; 
 prev = merge $ i; 
 let i = i + 1 
 done 
  

fact ls 列出排序中具有相同前缀的文件。如果您具有不同前缀的文件名，并且具有要输出它们的特定顺序，则可以修改它。

Hello I want to iteratively merge the files in input_directory and put the merged ones in output_directory.

Suppose in input_directory I have : file1.txt, file2.txt, file3.txt

I want to the output directory to contain the following files:

• merge1.txt : same as file1.txt
• merge2.txt : merge file1.txt file2.txt
• merge3.txt : merge file1.txt file2.txt file3.txt

I have almost no experience with bash scripts, it is obvious that it can be done with an iterator in a for loop, but I don't know how.

Thank you in advance..

解决方案

Store the previous filename that was appended to output and use it during next iteration:

#!/bin/bash

touch prev #a temp variable to hold previous filename
i=1;

for file in $(ls /INPUT/file*); 
do 
   cat /OUTPUT/$prev /INPUT/$file > /OUTPUT/merge$i;
   prev=merge$i;
   let i=i+1
 done

This code takes advantage of the fact ls list the files in sorted when they have same prefix. You can modify it if you have filenames with different prefixes and have a specific order in which you want to output them.

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