bash sh：测试文件是否存在
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sh: Test for existence of files
提问by ikegami
How does one test for the existence of files in a directory using bash?
如何使用 测试目录中文件的存在bash？
if ... ; then
echo 'Found some!'
fi
To be clear, I don't want to test for the existence of a specificfile. I would like to test if a specific directory contains anyfiles.
明确地说，我不想测试特定文件的存在。我想测试特定目录是否包含任何文件。
I went with:
我和：
(
shopt -s dotglob nullglob
existing_files=( ./* )
if [[ ${#existing_files[@]} -gt 0 ]] ; then
some_command "${existing_files[@]}"
fi
)
Using the array avoids race conditions from reading the file list twice.
使用数组可以避免竞争条件读取文件列表两次。
回答by Conrad Shultz
From the man page:
从手册页：
-f file
True if file exists and is a regular file.
So:
所以：
if [ -f someFileName ]; then echo 'Found some!'; fi
Edit: I see you already got the answer, but for completeness, you can use the info in Checking from shell script if a directory contains files- and lose the dotglob option if you want hidden files ignored.
编辑：我看到你已经得到了答案，但为了完整起见，如果目录包含文件，你可以使用从 shell 脚本检查中的信息- 如果你想忽略隐藏文件，则丢失 dotglob 选项。
回答by Conrad Shultz
I typically just use a cheap ls -A to see if there's a response.
我通常只使用便宜的 ls -A 来查看是否有响应。
Pseudo-maybe-correct-syntax-example-ahoy:
伪可能是正确的语法示例啊：
if [[ $(ls -A my_directory_path_variable ) ]] then....
edit, this will work:
编辑，这将工作：
myDir=(./*) if [ ${#myDir[@]} -gt 1 ]; then echo "there's something down here"; fi
myDir=(./*) if [ ${#myDir[@]} -gt 1 ]; then echo "there's something down here"; fi
回答by paxdiablo
You can use lsin an ifstatement thus:
您可以ls在if语句中使用：
if [[ "$(ls -a1 | egrep -v '^\.$|^\.\.$')" = "" ]] ; then echo empty ; fi
or, thanks to ikegami,
或者，感谢池上，
if [[ "$(ls -A)" = "" ]] ; then echo empty ; fi
or, even shorter:
或者，甚至更短：
if [[ -z "$(ls -A)" ]] ; then echo empty ; fi
These basically list all files in the current directory (including hidden ones) that are neither .nor ...
这些基本上列出了当前目录中的所有文件（包括隐藏的），既不是也不.是...
If that list is empty, then the directory is empty.
如果该列表为空，则该目录为空。
If you want to discount hidden files, you can simplify it to:
如果要打折隐藏文件，可以简化为：
if [[ "$(ls)" = "" ]] ; then echo empty ; fi
A bash-only solution (no invoking external programs like lsor egrep) can be done as follows:
一个bash-only 解决方案（不调用外部程序，如ls或egrep）可以按如下方式完成：
emp=Y; for i in *; do if [[ $i != "*" ]]; then emp=N; break; fi; done; echo $emp
It's not the prettiestcode in the world, it simply sets empto Yand then, for every real file, sets it to Nand breaks from the forloop for efficiency. If there were zero files, it stays as Y.
它不是世界上最漂亮的代码，它只是设置emp为Y，然后对于每个真实文件，将其设置为N并中断for循环以提高效率。如果有零个文件，它保持为Y.
回答by Steve Robillard
Try this
尝试这个
if [ -f /tmp/foo.txt ]
then
echo the file exists
fi
ref: http://tldp.org/LDP/abs/html/fto.html
参考：http: //tldp.org/LDP/abs/html/fto.html
you may also want to check this out: http://tldp.org/LDP/abs/html/fto.html
你可能还想看看这个：http: //tldp.org/LDP/abs/html/fto.html
How about this for whether directory is empty or not
这个目录是否为空怎么样
$ find "/tmp" -type f -exec echo Found file {} \;
回答by Xaxxon
#!/bin/bash
if [ -e ]; then
echo "File exists"
else
echo "Files does not exist"
fi
回答by Keith Thompson
I don't have a good pure sh/bash solution, but it's easy to do in Perl:
我没有好的纯 sh/bash 解决方案，但在 Perl 中很容易做到：
#!/usr/bin/perl
use strict;
use warnings;
die "Usage: if emptydir dir ; then
echo "dir is empty"
else
echo "dir is not empty"
fi
dir\n" if scalar @ARGV != 1 or not -d $ARGV[0];
opendir my $DIR, $ARGV[0] or die "$ARGV[0]: $!\n";
my @files = readdir $DIR;
closedir $DIR;
if (scalar @files == 2) { # . and ..
exit 0;
}
else {
exit 1;
}
Call it something like emptydirand put it somewhere in your $PATH, then:
emptydir将其命名为类似并将其放在您的某个位置$PATH，然后：
# tested on Linux BASH
directory=
if test $(stat -c %h $directory) -gt 2;
then
echo "not empty"
else
echo "empty"
fi
It dies with an error message if you give it no arguments, two or more arguments, or an argument that isn't a directory; it's easy enough to change if you prefer different behavior.
如果你不给它任何参数、两个或更多参数，或者一个不是目录的参数，它就会死，并显示一条错误消息；如果您喜欢不同的行为，那么改变就很容易了。
回答by Vidul
if ( shopt -s nullglob ; perl -e'exit !@ARGV' ./* ) ; then
echo 'Found some!'
fi
回答by ikegami
For fun:
为了娱乐：
##代码##(Doesn't check for hidden files)
（不检查隐藏文件）
