基于菜单的shell脚本删除文件

Question

我的要求是根据显示的内容删除文件。以下是我列出文件的代码片段，但是当我选择选项时它会显示文件，当我捕获文件名时，它不会发生，只有获取键不是VALUE（$ REPLY只显示键但不显示值）。有人可以帮助我。

#!/bin/bash
    select list in $(ls *.tmp)
    do
            echo  $list
            echo Do you want to delete files ?
            read userInput
            echo "UserInput is :: "  $userInput
            echo "Reply is :: " $REPLY
            if [ $userInput == $REPLY ] ; then
            #       rm $REPLY
                    echo 'Yes'
                    break
    done

---- ----- OUTPUT

1) +~JF1905393034413613060.tmp
 2) +~JF2032005334435574091.tmp
 3) +~JF3116454937363220082.tmp
 4) +~JF3334986634800781310.tmp
 5) +~JF3651229840772890748.tmp
 6) +~JF3882306323060007639.tmp
 7) +~JF573641658479505435.tmp
 8) +~JF6137053351660236007.tmp
 9) +~JF6277682393160684532.tmp
10) +~JF6385610668752278364.tmp
11) +~JF6824954027739238354.tmp
12) +~JF7876557427734797684.tmp
#? 4
+~JF3334986634800781310.tmp
Do you want to delete files ?
y
UserInput is ::  y
Reply is ::  4
No

Answer 1

试试这个：

PS3="Pick a file number to delete (or Ctrl-C to quit): "
select f in *.tmp ; do echo rm "$f" ; done 

然后删除echo以使其实际删除文件。