一举测试多个文件条件（BASH）？
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Test multiple file conditions in one swoop (BASH)?
提问by Alex Gray
Often when writing for the bash shell, one needs to test if a file (or Directory) exists (or doesn't exist) and take appropriate action. Most common amongst these test are...
通常在为 bash shell 编写代码时，需要测试文件（或目录）是否存在（或不存在）并采取适当的措施。这些测试中最常见的是...
-e- file exists, -f- file is a regular file (not a directory or device file), -s- file is not zero size, -d- file is a directory, -r- file has read permission, -w- file has write, or -xexecute permission (for the user running the test)
-e- 文件存在，-f- 文件是一个普通文件（不是目录或设备文件），-s- 文件不是零大小，-d- 文件是一个目录，-r- 文件有读权限，-w- 文件有写或-x执行权限（对于用户运行测试）
This is easily confirmed as demonstrated on this user-writable directory....
正如此用户可写目录中所示，这很容易确认....
#/bin/bash
if [ -f "/Library/Application Support" ]; then
echo 'YES SIR -f is fine'
else echo 'no -f for you'
fi
if [ -w "/Library/Application Support" ]; then
echo 'YES SIR -w is fine'
else echo 'no -w for you'
fi
if [ -d "/Library/Application Support" ]; then
echo 'YES SIR -d is fine'
else echo 'no -d for you'
fi
? no -f for you ?
? YES SIR -w is fine ?
? YES SIR -d is fine ?
? 没有 -f 吗？
? 是的，先生 -w 很好吗？
? 是的 SIR -d 可以吗？
My question, although seemingly obvious, and unlikely to be impossible - is how to simply combinethese tests, without having to perform them separately for each condition... Unfortunately...
我的问题虽然看起来很明显，也不太可能是不可能的 - 是如何简单地组合这些测试，而不必为每个条件单独执行它们......不幸的是......
if [ -wd "/Library/Application Support" ]
? -wd: unary operator expected
if [ -w | -d "/Library/Application Support" ]
? [: missing `]'
? -d: command not found
if [ -w [ -d "/Library.... ]] & if [ -w && -d "/Library.... ]
? [: missing `]'
? no -wd for you ?
? no -w | -d for you ?
? no [ -w [ -d .. ]] for you ?
? no -w && -d for you ?
? 没有 -wd 吗？
? 没有 -w | -d 给你？
? 没有 [ -w [ -d .. ]] 给你？
? 没有 -w && -d 吗？
What am I missing here?
我在这里缺少什么？
回答by Kerrek SB
You can use logical operators to multiple conditions, e.g. -afor AND:
您可以使用逻辑运算符将多个条件，例如-a为AND：
MYFILE=/tmp/data.bin
if [ -f "$MYFILE" -a -r "$MYFILE" -a -w "$MYFILE" ]; then
#do stuff
fi
unset MYFILE
回答by Micha? ?rajer
Of course, you need to use AND somehow as Kerrek(+1) and Ben(+1) pointed it out. You can do in in few different ways. Here is an ala-microbenchmark results for few methods:
当然，正如 Kerrek(+1) 和 Ben(+1) 指出的那样，您需要以某种方式使用 AND。您可以通过几种不同的方式进行。以下是几种方法的 ala-microbenchmark 结果：
Most portable and readable way:
最便携和可读的方式：
$ time for i in $(seq 100000); do [ 1 = 1 ] && [ 2 = 2 ] && [ 3 = 3 ]; done
real 0m2.583s
still portable, less readable, faster:
仍然便携，可读性差，速度更快：
$ time for i in $(seq 100000); do [ 1 = 1 -a 2 = 2 -a 3 = 3 ]; done
real 0m1.681s
bashism, but readable and faster
bashism，但可读性更强，速度更快
$ time for i in $(seq 100000); do [[ 1 = 1 ]] && [[ 2 = 2 ]] && [[ 3 = 3 ]]; done
real 0m1.285s
bashism, but quite readable, and fastest.
bashism，但相当可读，而且速度最快。
$ time for i in $(seq 100000); do [[ 1 = 1 && 2 = 2 && 3 = 3 ]]; done
real 0m0.934s
Note, that in bash, "[" is a builtin, so bash is using internal command not a symlink to /usr/bin/test exacutable. The "[[" is a bash keyword. So the slowest possible way will be:
请注意，在 bash 中，“[”是一个内置命令，因此 bash 使用的是内部命令，而不是指向 /usr/bin/test 可执行文件的符号链接。"[[" 是一个 bash 关键字。所以最慢的方法是：
time for i in $(seq 100000); do /usr/bin/\[ 1 = 1 ] && /usr/bin/\[ 2 = 2 ] && /usr/bin/\[ 3 = 3 ]; done
real 14m8.678s
回答by Ben Hymanson
You want -aas in -f foo -a -d foo(actually that test would be false, but you get the idea).
你想-a在-f foo -a -d foo（实际上是测试是假的，但你的想法）。
You were close with &you just needed &&as in [ -f foo ] && [ -d foo ]although that runs multiple commands rather than one.
尽管它运行多个命令而不是一个命令，但您与&您的需求很接近。&&[ -f foo ] && [ -d foo ]
Here is a manual page for testwhich is the command that [is a link to. Modern implementations of testhave a lot more features (along with the shell-builtin version [[which is documented in your shell's manpage).
这是一个用于测试的手册页，它[是一个链接到的命令。的现代实现test具有更多功能（以及[[shell 联机帮助页中记录的 shell 内置版本）。
回答by jonretting
check-file(){
while [[ ${#} -gt 0 ]]; do
case in
fxrsw) [[ -f "" && -x "" && -r "" && -s "" && -w "" ]] || return 1 ;;
fxrs) [[ -f "" && -x "" && -r "" && -s "" ]] || return 1 ;;
fxr) [[ -f "" && -x "" && -r "" ]] || return 1 ;;
fr) [[ -f "" && -r "" ]] || return 1 ;;
fx) [[ -f "" && -x "" ]] || return 1 ;;
fe) [[ -f "" && -e "" ]] || return 1 ;;
hf) [[ -h "" && -f "" ]] || return 1 ;;
*) [[ -e "" ]] || return 1 ;;
esac
shift
done
}
check-file fxr "/path/file" && echo "is valid"
check-file hf "/path/folder/symlink" || { echo "Fatal error cant validate symlink"; exit 1; }
check-file fe "file.txt" || touch "file.txt" && ln -s "${HOME}/file.txt" "/docs/file.txt" && check-file hf "/docs/file.txt" || exit 1
if check-file fxrsw "${HOME}"; then
echo "Your home is your home from the looks of it."
else
echo "You infected your own home."
fi
回答by chad
Why not write a function to do it?
为什么不写一个函数来做呢？
check_file () {
local FLAGS=
local PATH=
if [ -z "$PATH" ] ; then
if [ -z "$FLAGS" ] ; then
echo "check_file: must specify at least a path" >&2
exit 1
fi
PATH=$FLAGS
FLAGS=-e
fi
FLAGS=${FLAGS#-}
while [ -n "$FLAGS" ] ; do
local FLAG=`printf "%c" "$FLAGS"`
if [ ! -$FLAG $PATH ] ; then false; return; fi
FLAGS=${FLAGS#?}
done
true
}
Then just use it like:
然后像这样使用它：
for path in / /etc /etc/passwd /bin/bash
{
if check_file -dx $path ; then
echo "$path is a directory and executable"
else
echo "$path is not a directory or not executable"
fi
}
And you should get:
你应该得到：
/ is a directory and executable
/etc is a directory and executable
/etc/passwd is not a directory or not executable
/bin/bash is not a directory or not executable
回答by Gao
This seems to work (notice the double brackets):
这似乎有效（注意双括号）：
#!/bin/bash
if [[ -fwd "/Library/Application Support" ]]
then
echo 'YES SIR -f -w -d are fine'
else
echo 'no -f or -w or -d for you'
fi
