是否有一个bash命令来计算匹配模式的文件数量?
例如,我想获取目录中所有文件的计数,这些文件都符合这个模式:log*
是否有一个bash命令来计算匹配模式的文件数量?
例如,我想获取目录中所有文件的计数,这些文件都符合这个模式:log*
当前回答
这个简单的一行代码可以在任何shell中工作,而不仅仅是bash:
ls -1q log* | wc -l
Ls -1q将为每个文件提供一行,即使它们包含空格或换行符等特殊字符。
输出被输送到wc -l,它计算行数。
其他回答
你可以用bash安全地做到这一点(即不会被名称中有空格或\n的文件所bug):
$ shopt -s nullglob
$ logfiles=(*.log)
$ echo ${#logfiles[@]}
您需要启用nullglob,以便在没有匹配的文件时不会在$logfiles数组中获得*.log文本。(参见如何“撤消”一个“set -x”?有关如何安全重置的示例。)
这是我的一句话。
file_count=$( shopt -s nullglob ; set -- $directory_to_search_inside/* ; echo $#)
这可以用标准POSIX shell语法完成。
下面是一个简单的count_entries函数:
#!/usr/bin/env sh
count_entries()
{
# Emulating Bash nullglob
# If argument 1 is not an existing entry
if [ ! -e "$1" ]
# argument is a returned pattern
# then shift it out
then shift
fi
echo $#
}
对于紧凑的定义:
count_entries(){ [ ! -e "$1" ]&&shift;echo $#;}
特色POSIX兼容的文件计数器类型:
#!/usr/bin/env sh
count_files()
# Count the file arguments matching the file operator
# Synopsys:
# count_files operator FILE [...]
# Arguments:
# $1: The file operator
# Allowed values:
# -a FILE True if file exists.
# -b FILE True if file is block special.
# -c FILE True if file is character special.
# -d FILE True if file is a directory.
# -e FILE True if file exists.
# -f FILE True if file exists and is a regular file.
# -g FILE True if file is set-group-id.
# -h FILE True if file is a symbolic link.
# -L FILE True if file is a symbolic link.
# -k FILE True if file has its `sticky' bit set.
# -p FILE True if file is a named pipe.
# -r FILE True if file is readable by you.
# -s FILE True if file exists and is not empty.
# -S FILE True if file is a socket.
# -t FD True if FD is opened on a terminal.
# -u FILE True if the file is set-user-id.
# -w FILE True if the file is writable by you.
# -x FILE True if the file is executable by you.
# -O FILE True if the file is effectively owned by you.
# -G FILE True if the file is effectively owned by your group.
# -N FILE True if the file has been modified since it was last read.
# $@: The files arguments
# Output:
# The number of matching files
# Return:
# 1: Unknown file operator
{
operator=$1
shift
case $operator in
-[abcdefghLkprsStuwxOGN])
for arg; do
# If file is not of required type
if ! test "$operator" "$arg"; then
# Shift it out
shift
fi
done
echo $#
;;
*)
printf 'Invalid file operator: %s\n' "$operator" >&2
return 1
;;
esac
}
count_files "$@"
示例用法:
count_files -f log*.txt
count_files -d datadir*
计数没有循环的非目录条目:
#!/bin/sh
# Creates strings of as many dots as expanded arguments
# dotted string for entries matching star pattern
star=$(printf '%.0s.' ./*)
# dotted string for entries matching star slash pattern (directories)
star_dir=$(printf '%.0s.' ./*/)
# dotted string for entries matching dot star pattern
dot_star=$(printf '%.0s.' ./.*)
# dotted string for entries matching dot star slash pattern (directories)
dot_star_dir=$(printf '%.0s.' ./.*/)
# Print pattern matches count excluding directories matches
printf 'Files count: %d\n' $((
${#star} - ${#star_dir} +
${#dot_star} - ${#dot_star_dir}
))
如果您有很多文件,并且不想使用优雅的shopt -s nullglob和bash数组解决方案,您可以使用find等,只要不打印文件名(可能包含换行符)。
find -maxdepth 1 -name "log*" -not -name ".*" -printf '%i\n' | wc -l
这将找到所有匹配log*且不以。*开头的文件-“not name .*”是冗余的,但重要的是要注意,“ls”的默认值是不显示点文件,但find的默认值是包含它们。
这是一个正确答案,可以处理任何类型的文件名,因为文件名永远不会在命令之间传递。
但是,shopt nullglob答案是最好的答案!
这个简单的一行代码可以在任何shell中工作,而不仅仅是bash:
ls -1q log* | wc -l
Ls -1q将为每个文件提供一行,即使它们包含空格或换行符等特殊字符。
输出被输送到wc -l,它计算行数。