Bash if [ -d $1] 为空 $1 返回 true
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/21678771/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Bash if [ -d $1] returning true for empty $1
提问by fgysin reinstate Monica
So I have the following little script and keep wondering..
所以我有以下小脚本并不断想知道..
#!/bin/bash
if [ -d ]; then
echo 'foo'
else
echo 'bar'
fi
.. why does this print foo when called without arguments? How is it that the test [-d ] returns true for an empty string?
.. 为什么在不带参数的情况下调用时会打印 foo ？测试 [-d ] 如何为空字符串返回 true？
采纳答案by fedorqui 'SO stop harming'
From: info coreutils 'test invocation'(reference found through man test):
来自：（info coreutils 'test invocation'通过 找到的参考man test）：
If EXPRESSION is omitted,
test' returns false. **If EXPRESSION is a single argument,test' returns false if the argument is null and true otherwise**. The argument can be any string, including strings like-d',-1',--',--help', and--version' that most other programs would treat as options. To get help and version information, invoke the commands[ --help' and `[ --version', without the usual closing brackets.
如果省略 EXPRESSION，
test' returns false. **If EXPRESSION is a single argument,则如果参数为 null ，则test' 返回 false，否则返回 true**。参数可以是任何字符串，包括像-d',-1'、----',help' 和--version' that most other programs would treat as options. To get help and version information, invoke the commands[ --help' 和 `[ --version'这样的字符串 ，没有通常的右括号。
Highlighting properly:
正确突出显示：
If EXPRESSION is a single argument, `test' returns false if the argument is null and true otherwise
如果 EXPRESSION 是单个参数，如果参数为 null，`test' 返回 false，否则返回 true
So whenever we do [ something ]it will return trueif that somethingis not null:
所以每当我们这样做时[ something ]，true如果它something不为空，它就会返回：
$ [ -d ] && echo "yes"
yes
$ [ -d "" ] && echo "yes"
$
$ [ -f ] && echo "yes"
yes
$ [ t ] && echo "yes"
yes
Seeing the second one [ -d "" ] && echo "yes"returning false, you get the way to solve this issue: quote $1so that -dalways gets a parameter:
看到第二个[ -d "" ] && echo "yes"返回false，你就有了解决这个问题的方法：quote$1这样-d总是得到一个参数：
if [ -d "" ]; then
echo 'foo'
else
echo 'bar'
fi
回答by devnull
The reason that
原因是
[ -d ] && echo y
produces yis that the shell interprets it as a stringin the testcommand and evaluates it to true. Even saying:
产生y的是，外壳将其解释为一个字符串的test命令，并将其评估为TRUE。甚至说：
[ a ] && echo y
would produce y. Quoting from help test:
会产生y. 引自help test：
string True if string is not the null string.
That is why quoting variables is recommended. Saying:
这就是为什么推荐引用变量的原因。说：
[ -d "" ] && echo y
should not produce ywhen called without arguments.
不y带参数调用时不应产生。
回答by Alfe
The reason is plain and simple: The syntax does not match the case in which the -dis recognized as an operator working on a file name. It is just taken as a string, and each non-empty string is true. Only if a second parameter to -dis given, it is recognized as the operator to find out whether a given FILE is a directory.
原因很简单：语法与将-d识别为处理文件名的运算符的情况不匹配。它只是作为一个字符串，每个非空字符串都为真。只有-d给定了第二个参数to ，才被认为是判断给定FILE是否为目录的运算符。
The same applies to all the other operators like -e, -r, etc.
这同样适用于所有其他运营商一样-e，-r等等。
In your case, use double quotes to avoid running into that "problem":
在您的情况下，请使用双引号以避免遇到该“问题”：
[ -d "" ]
