如何使用变量作为查找命令的路径 [英] How to use variable as a path for find command

问题描述

可以有人给我一个提示，我可以如何使下面的bash脚本正常运行：

  path =/ path /
 excludepath =！-path \/ path / to / exclude / * \
 
找到$ path$ excludepath-name *。txt -type f 
  

当我尝试运行它时，我得到：

  find：`！ -path/ path / to / exclude / *'：没有这样的文件或目录
  

谢谢您的问题是事实上，$ excludepath >是一个参数，即使它包含空格。要放置几个参数，请使用字符串数组而不是字符串：

  excludepath =（！-path* webshow * ）
找到$ path$ {excludepath [@]}-name* .txt-type f 
  

这是一个bashism，但是（绊倒这个的人不要使用 bash 将需要做其他事情）

Can someone give me a hint, how I can make the following bash script run properly:

path="/path/"
excludepath="! -path \"/path/to/exclude/*\" "

find "$path" "$excludepath" -name *."txt" -type f

When I try to run it I get:

find: `! -path "/path/to/exclude/*"': No such file or directory

Thank you in advance.

解决方案

Your problem was the fact that "$excludepath" is a single argument, even if it contains spaces. To put several arguments, use an array of strings instead of a string:

excludepath=(! -path "*webshow*")
find "$path" "${excludepath[@]}" -name "*.txt" -type f

This is a bashism, though (people that stumble onto this and don't use bash will need to do something else).

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