Question

我想使用shell脚本列出所有目录。我使用以下代码：

DIR="$1"

if [ $# -ne 1 ]
then
    echo "Usage: $0 {dir-name}"
exit 1
fi

cd "$DIR"
SAVEIFS=$IFS
IFS=$(echo -en "\n\b")
for user in $( ls -d */)
do
  for dirs in $( ls -d $PWD/$user*)
  do
    echo $PWD/$user/$dirs;
  done
done
IFS=$SAVEIFS

如果该目录上没有任何空格，它对我有用，否则它会为其上的每个空格分割输出。我得到了以下输出：

abhinaba@abhinaba-desktop:~/software$ sh test.sh /media/2C44138344134F48/RB1
ls: cannot access /m: No such file or directory
ls: cannot access dia/2C44138344134F48/RB1/VB*: No such file or directory
ls: cannot access /m: No such file or directory
ls: cannot access dia/2C44138344134F48/RB1/DLI*: No such file or directory
ls: cannot access /m: No such file or directory
ls: cannot access dia/2C44138344134F48/RB1/3001/*: No such file or directory
ls: cannot access /m: No such file or directory
ls: cannot access dia/2C44138344134F48/RB1/VB*: No such file or directory
ls: cannot access /m: No such file or directory
ls: cannot access dia/2C44138344134F48/RB1/DLI*: No such file or directory
ls: cannot access /m: No such file or directory
ls: cannot access dia/2C44138344134F48/RB1/3002/*: No such file or directory
ls: cannot access /m: No such file or directory
ls: cannot access dia/2C44138344134F48/RB1/VB*: No such file or directory
ls: cannot access /m: No such file or directory
ls: cannot access dia/2C44138344134F48/RB1/DLI*: No such file or directory
ls: cannot access /m: No such file or directory
ls: cannot access dia/2C44138344134F48/RB1/3003/*: No such file or directory

Answer 1

已编辑：已移除ls以优化脚本（来自tripleee）

如果你使用BASH shell：

#!/bin/bash

SAVEIFS=$IFS
IFS=$(echo -en "\n\b")
DIR="$1"

if [ $# -ne 1 ]
 then
 echo "Usage: $0 {dir-name}"
 exit 1
fi

cd "$DIR"

for user in $(ls -d */)
 do
 for dirs in $(ls -d $user*)
  do
  echo $dirs
 done
done

IFS=$SAVEIFS

运行上面的“test.sh”脚本，如下所示：

bash test.sh /media/2C44138344134F48/RB1

或简单地（如果您已经在BASH中并且已设置了eXecutable标志）

./test.sh /media/2C44138344134F48/RB1

Reference

Answer 2

使用find实际上最简单：

find PATH -type d -name '* *'

或者，如果您需要对每个结果执行某些操作，请考虑将其连接到xargs

find PATH -type d -name '* *' -print0 | xargs -0 run-some-command

或者，如果您只是希望所有目录安全地转义为参数，那么：

find PATH -type d -print0 | xargs -0 run-some-command

在run-some-command内，脚本的每个参数都将正确设置为每个目录名称，无论它包含哪些字符。

Answer 3

下面的简单脚本将满足您的需求。

#!/bin/bash

function usage() {
        echo "Usage: `basename $0` <dir-name>"
}


if [ "x$1" = "x" ]
then
        usage
        exit 0
fi

find "$1" -type 'd'

列出名称中包含空格的所有目录

3 个答案: