bash 使用 [! -d "$DIR"]
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Command not found in Bash's IF-ELSE condition when using [! -d "$DIR"]
提问by neversaint
I have a code like this
我有这样的代码
#!/bin/bash
DIR="test_dir/";
if [! -d "$DIR"]; then
# If it doesn't create it
mkdir $DIR
fi
But why executing it gave me this:
但是为什么执行它给了我这个：
./mycode.sh: line 16: [!: command not found
What's the right way to do it?
正确的做法是什么？
回答by konsolebox
Add space between [ and !. And before ] as well.
在 [ 和 ! 之间添加空格。在 ] 之前也是如此。
#!/bin/bash
DIR="test_dir/";
if [ ! -d "$DIR" ]; then
# If it doesn't create it
mkdir $DIR
fi
It's also a good idea to quote your variable:
引用您的变量也是一个好主意：
mkdir "$DIR"
回答by falsetru
Add some spaces:
添加一些空格：
if [ ! -d "$DIR" ]; then
# ^ ^
回答by devnull
You could also attempt to simply by saying:
您也可以尝试简单地说：
test -d "${dir}" || mkdir "${dir}"
This would create the directory if it doesn't exist.
如果目录不存在，这将创建该目录。
