查找/排序解决方案工作得很好，直到文件数量变得非常大(比如整个文件系统)。使用awk来跟踪最近的文件:

find $DIR -type f -printf "%T@ %p\n" | 
awk '
BEGIN { recent = 0; file = "" }
{
if ($1 > recent)
   {
   recent = $1;
   file = $0;
   }
}
END { print file; }' |
sed 's/^[0-9]*\.[0-9]* //'

这是一个递归版本(即，它在某个目录或其任何子目录中查找最近更新的文件)

find /dir/path -type f -printf "%T@ %p\n" | sort -n | cut -d' ' -f 2- | tail -n 1

命令行简单的外行解释:

find /dir/path -type f finds all the files in the directory -printf "%T@ %p\n" prints a line for each file where %T@ is the float seconds since 1970 epoch and %p is the filename path and \n is the new line character for more info see man find | is a shell pipe (see man bash section on Pipelines) sort -n means to sort on the first column and to treat the token as numerical instead of lexicographic (see man sort) cut -d' ' -f 2- means to split each line using the character and then to print all tokens starting at the second token (see man cut) NOTE: -f 2 would print only the second token tail -n 1 means to print the last line (see man tail)

我喜欢echo *(om[1]) (zsh语法)，因为它只给出文件名，而不调用任何其他命令。

ls -lAtr | tail -1

其他解决方案不包括以“。”开头的文件。

该命令还包括'。'和'..，这可能是也可能不是你想要的:

Ls -latr | tail -1

基于dmckee回答的简短变体:

ls -t | head -1

只有Bash内置程序，紧跟BashFAQ/003:

shopt -s nullglob

for f in * .*; do
    [[ -d $f ]] && continue
    [[ $f -nt $latest ]] && latest=$f
done

printf '%s\n' "$latest"