在 if 语句 (Bash) 中执行代码
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Executing code in if-statement (Bash)
提问by Tyilo
New question:
新问题：
I can't do this (Error: line 2: [: ==: unary operator expected):
我不能这样做（错误：）line 2: [: ==: unary operator expected：
if [ $(echo "") == "" ]
then
echo "Success!"
fi
But this works fine:
但这工作正常：
tmp=$(echo "")
if [ "$tmp" == "" ]
then
echo "Success!"
fi
Why?
为什么？
Original question:
原问题：
Is it possible to get the result of a command inside an if-statement?
是否可以在 if 语句中获取命令的结果？
I want to do something like this:
我想做这样的事情：
if [ $(echo "foo") == "foo" ]
then
echo "Success!"
fi
I currently use this work-around:
我目前使用这个解决方法：
tmp=$(echo "foo")
if [ "$tmp" == "foo" ]
then
echo "Success!"
fi
回答by Chen Levy
The short answer is yes-- You can evaluate a command inside an ifcondition. The only thing I would change in your first example is the quoting:
简短的回答是肯定的——您可以在if条件内评估命令。在你的第一个例子中，我唯一要改变的是引用：
if [ "$(echo foo)" == "foo" ]
then
echo "Success"'!'
fi
- Note the funny quote for the
'!'. This disables the special behavior of!inside an interactive bash session, that might produce unexpected results for you.
- 请注意
'!'. 这会禁用!交互式 bash 会话内部的特殊行为，这可能会给您带来意想不到的结果。
After your update your problem becomes clear, and the change in quoting actually solves it:
更新后，您的问题变得清晰，引用中的更改实际上解决了它：
The evaluation of $(...)occurs before the evaluation of if [...], thus if $(...)evaluates to an empty string the [...]becomes if [ == ""]which is illegal syntax.
的评估$(...)发生在 的评估之前if [...]，因此如果$(...)评估为空字符串，则[...]成为if [ == ""]非法语法。
The way to solve this is having the quotes outside the $(...)expression. This is where you might get into the sticky issue of quoting inside quoting, but I will live this issue to another question.
解决这个问题的方法是在$(...)表达式之外加上引号。这是您可能会遇到内部引用引用的棘手问题的地方，但我会将这个问题解决到另一个问题。
