我可以省略的部分,然后在if语句? [英] Can I omit the then part in an if statement?
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问题描述
如何将是正确的bash语法是这样的:
How would be the correct bash syntax for something like this:
if [ "$actual" == "$expected" ]; then
doNothing
else
echo "Error: actual: $actual. Expected: $expected"
fi
我要寻找的东西,对于变量的实际和预期的所有可能值的作品。变量的含量应不PTED间$ P $ /评估/以任何方式扩大。该脚本并不需要移植(一个bash唯一的解决办法是确定)。
I am looking for something that works for all possible values of the variables "actual" and "expected". The content of the variables must not be interpreted/evaluated/expanded in any way. The script does not need to be portable (a bash only solution is ok).
推荐答案
您可以用最简单的可用什么也不做的语句:
You could use the simplest do-nothing statement available:
if [ "$actual" = "$expected" ]; then
:
else
echo "Error: actual: $actual. Expected: $expected"
fi
(注意:一个 =
不是两个 [
/ 测试
)
但是,一个更好的主意是刚刚反转测试并删除需要一种完全
But a better idea is to just invert the test and remove the need for that entirely:
if [ "$actual" != "$expected" ]; then
echo "Error: actual: $actual. Expected: $expected"
fi
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