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。

ls -1 log* | wc -l

这意味着每行列出一个文件，然后将其输送到单词计数命令，参数切换到计数行。

我已经对这个答案进行了很多思考，特别是考虑到不要解析这些东西。起初，我尝试过

<WARNING! DID NOT WORK>

du --inodes --files0-from=<(find . -maxdepth 1 -type f -print0) | awk '{sum+=int($1)}END{print sum}'

</WARNING! DID NOT WORK>

如果只有像这样的文件名，这是可行的

touch $'w\nlf.aa'

但如果我创建这样的文件名就失败了

touch $'firstline\n3 and some other\n1\n2\texciting\n86stuff.jpg'

我终于想到了下面这些。注意，我试图获得目录中所有文件的计数(不包括任何子目录)。我认为它，连同@Mat和@Dan_Yard的答案，以及至少有@mogsie设定的大部分要求(我不确定内存)。我认为@mogsie的答案是正确的，但我总是尽量避免解析ls，除非是非常特定的情况。

awk -F"\0" '{print NF-1}' < <(find . -maxdepth 1 -type f -print0) | awk '{sum+=$1}END{print sum}'

更可读的:

awk -F"\0" '{print NF-1}' < \
  <(find . -maxdepth 1 -type f -print0) | \
    awk '{sum+=$1}END{print sum}'

这是专门为文件执行查找，用空字符分隔输出(以避免空格和换行问题)，然后计算空字符的数量。文件的数量将比空字符的数量少一个，因为在最后会有一个空字符。

要回答OP的问题，有两种情况需要考虑

1)非递归搜索:

awk -F"\0" '{print NF-1}' < \
  <(find . -maxdepth 1 -type f -name "log*" -print0) | \
    awk '{sum+=$1}END{print sum}'

2)递归搜索。请注意，name参数中的内容可能需要针对略有不同的行为(隐藏文件等)进行更改。

awk -F"\0" '{print NF-1}' < \
  <(find . -type f -name "log*" -print0) | \
    awk '{sum+=$1}END{print sum}'

如果有人想评论这些答案与我在这个答案中提到的答案相比如何，请评论。

注意，我是在得到这个答案的同时进行这个思考过程的。

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}
))

这里有很多答案，但有些没有考虑在内

包含空格、换行符或控制字符的文件名 以连字符开头的文件名(想象一个名为-l的文件) 隐藏文件，以点开始(如果glob是*.log而不是log*) 匹配glob的目录(例如，一个名为logs的目录匹配log*) 空目录(即结果为0) 非常大的目录(列出所有目录会耗尽内存)

这里有一个解决方案可以处理所有这些问题:

ls 2>/dev/null -Ubad1 -- log* | wc -l

解释:

-U causes ls to not sort the entries, meaning it doesn't need to load the entire directory listing in memory -b prints C-style escapes for nongraphic characters, crucially causing newlines to be printed as \n. -a prints out all files, even hidden files (not strictly needed when the glob log* implies no hidden files) -d prints out directories without attempting to list the contents of the directory, which is what ls normally would do -1 makes sure that it's on one column (ls does this automatically when writing to a pipe, so it's not strictly necessary) 2>/dev/null redirects stderr so that if there are 0 log files, ignore the error message. (Note that shopt -s nullglob would cause ls to list the entire working directory instead.) wc -l consumes the directory listing as it's being generated, so the output of ls is never in memory at any point in time. -- File names are separated from the command using -- so as not to be understood as arguments to ls (in case log* is removed)

shell会将log*扩展到完整的文件列表，如果文件很多，可能会耗尽内存，所以通过grep运行会更好:

ls -Uba1 | grep ^log | wc -l

最后一种方法在不使用大量内存的情况下处理超大文件目录(尽管它使用了子shell)。不再需要-d，因为它只列出当前目录的内容。

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。)