好的，这是我在Stack Overflow上的第一篇文章!

Though my problems with this have always been in csh not bash the solution I present will, I'm sure, work in both. The issue is with the shell's interpretation of the "ls" returns. We can remove "ls" from the problem by simply using the shell expansion of the * wildcard - but this gives a "no match" error if there are no files in the current (or specified folder) - to get around this we simply extend the expansion to include dot-files thus: * .* - this will always yield results since the files . and .. will always be present. So in csh we can use this construct ...

foreach file (* .*)
   echo $file
end

如果你想过滤掉标准的点文件，那很简单…

foreach file (* .*)
   if ("$file" == .) continue
   if ("file" == ..) continue
   echo $file
end

在这个线程的第一篇文章中的代码将这样写:-

getlist() {
  for f in $(* .*)
  do
    echo "File found: $f"
    # do something useful
  done
}

希望这能有所帮助!

好的，这是我在Stack Overflow上的第一篇文章!

Though my problems with this have always been in csh not bash the solution I present will, I'm sure, work in both. The issue is with the shell's interpretation of the "ls" returns. We can remove "ls" from the problem by simply using the shell expansion of the * wildcard - but this gives a "no match" error if there are no files in the current (or specified folder) - to get around this we simply extend the expansion to include dot-files thus: * .* - this will always yield results since the files . and .. will always be present. So in csh we can use this construct ...

foreach file (* .*)
   echo $file
end

如果你想过滤掉标准的点文件，那很简单…

foreach file (* .*)
   if ("$file" == .) continue
   if ("file" == ..) continue
   echo $file
end

在这个线程的第一篇文章中的代码将这样写:-

getlist() {
  for f in $(* .*)
  do
    echo "File found: $f"
    # do something useful
  done
}

希望这能有所帮助!

因为你没有使用find做任何其他类型的过滤，你可以使用以下bash 4.0:

shopt -s globstar
getlist() {
    for f in **/foo*
    do
        echo "File found: $f"
        # do something useful
    done
}

**/将匹配零个或多个目录，因此完整模式将匹配当前目录或任何子目录中的foo*。

find . -name "fo*" -print0 | xargs -0 ls -l

见xargs先生。

我真的很喜欢for循环和数组迭代，所以我想我会把这个答案添加到混合…

我也喜欢marchelbling愚蠢的文件示例。：）

$ mkdir test
$ cd test
$ touch "stupid file1"
$ touch "stupid file2"
$ touch "stupid   file 3"

在test目录中:

readarray -t arr <<< "`ls -A1`"

这将每个文件列表行添加到名为arr的bash数组中，并删除任何尾随换行符。

假设我们想给这些文件取一个更好的名字……

for i in ${!arr[@]}
do 
    newname=`echo "${arr[$i]}" | sed 's/stupid/smarter/; s/  */_/g'`; 
    mv "${arr[$i]}" "$newname"
done

$ {!arr[@]}扩展到0 1 2，因此"${arr[$i]}"是数组的第i个元素。变量周围的引号对于保留空格很重要。

结果是三个重命名的文件:

$ ls -1
smarter_file1
smarter_file2
smarter_file_3

我最近不得不处理一个类似的情况，我构建了一个FILES数组来遍历文件名:

eval FILES=($(find . -iname "foo*" -printf '"%p" '))

这里的想法是用双引号包围每个文件名，用空格分隔它们，并使用结果初始化FILES数组。 必须使用eval来正确计算find输出中的双引号，以初始化数组。

要遍历文件，只需执行:

for f in "${FILES[@]}"; do
    # Do something with $f
done