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通过 shell 脚本分割“文件”和“带空格的目录名”

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通过 shell 脚本分割“文件”和“带空格的目录名”

我有一个名为的文件，Files.txt其内容如下：

TestApp/Resources/Supporting Files/main.m
TestApp/Resources/Supporting Files/AppDelegate.h
TestApp/Resources/Supporting Files/AppDelegate.m

我正在提取文件和目录名称，如下所示，并将它们传递给另一个进程。

files=$(cat Files.txt)

for item in $files ; do    
  dn=$(dirname $item)

  printf $item
  printf "\n"
  printf $dn
  printf "\n\n"

  # passing to another process
done

然而，这让我这样：

TestApp/Resources/Supporting
TestApp/Resources

Files/main.m
Files

TestApp/Resources/Supporting
TestApp/Resources

Files/AppDelegate.h
Files

TestApp/Resources/Supporting
TestApp/Resources

Files/AppDelegate.m
Files

我需要的是这样的：

TestApp/Resources/Supporting Files/main.m
TestApp/Resources/Supporting Files

TestApp/Resources/Supporting Files/AppDelegate.h
TestApp/Resources/Supporting Files

TestApp/Resources/Supporting Files/AppDelegate.m
TestApp/Resources/Supporting Files

我尝试\在Files.txtas 中添加空格前缀：

TestApp/Resources/Supporting\ Files/main.m

并%20作为：

TestApp/Resources/Supporting%20Files/main.m

没有运气！

答案1

  1. for循环迭代字不是线条
  2. 总是引用你的"$variables"（除非你确切知道什么时候不引用）
while read -r item ; do    
  dn=$(dirname "$item")

  printf "%s\n" "$item"
  printf "%s\n" "$dn"

  # pass "$item" and "$dn" to another process
done < Files.txt

答案2

您需要设置字段分隔符：

OIFS=$IFS  
IFS=$'\n'

files=$(cat Files.txt)

for item in $files ; do    
  dn=$(dirname $item)
  printf $item
  printf "\n"
  printf $dn
  printf "\n\n"

  # passing to another process
done

IFS=$OIFS

输出：

[me@localhost test]$ ./test.sh 
TestApp/Resources/Supporting Files/main.m
TestApp/Resources/Supporting Files

TestApp/Resources/Supporting Files/AppDelegate.h
TestApp/Resources/Supporting Files

TestApp/Resources/Supporting Files/AppDelegate.m
TestApp/Resources/Supporting Files

解释： http://en.wikipedia.org/wiki/Internal_field_separator

该$IFS变量定义输入如何拆分为标记，默认为空格、制表符和换行符。由于您只想在换行符上拆分，因此$IFS需要临时更改该变量。

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