Bash 条件：如何“和”表达式？（如果 [ ! -z $VAR && -e $VAR ]）
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Bash conditionals: how to "and" expressions? (if [ ! -z $VAR && -e $VAR ])
提问by atxdba
I guess I'm not clear on how to do "and" tests. I wanted to make sure an argument existed which was working well with [ -e $VAR ], but it turns out that was also evaluating as true on an empty string; which I do not want.
我想我不清楚如何进行“和”测试。我想确保存在一个与 配合良好的参数[ -e $VAR ]，但事实证明，它在空字符串上也被评估为真；我不想要。
How do I 'and' them together? Or is there another unary test that accomplishes what I want?
我如何“和”他们在一起？或者是否有另一个一元测试可以完成我想要的？
回答by jaypal singh
if [ ! -z "$var" ] && [ -e "$var" ]; then
# something ...
fi
回答by paxdiablo
From the bashmanpage:
从bash联机帮助页：
[[ expression ]]- return a status of 0 or 1 depending on the evaluation of the conditional expression expression.
[[ expression ]]- 根据条件表达式表达式的计算结果返回 0 或 1 状态。
And, for expressions, one of the options is:
而且，对于表达式，选项之一是：
expression1 && expression2- true if bothexpression1andexpression2are true.
expression1 && expression2- 如果expression1和expression2都为真，则为真。
So you can andthem together as follows (-nis the opposite of -zso we can get rid of the !):
因此，您可以and按如下-n方式将它们组合在一起（与-z这样我们可以摆脱!）：
if [[ -n "$var" && -e "$var" ]] ; then
echo "'$var' is non-empty and the file exists"
fi
However, I don't think it's needed in this case, -e xyzzyis true if the xyzzyfileexists and can quite easily handle empty strings. If that's what you want then you don't actually need the -znon-empty check:
但是，我认为在这种情况下不需要它，-e xyzzy如果xyzzy文件存在并且可以很容易地处理空字符串，则为 true 。如果那是您想要的，那么您实际上并不需要-z非空检查：
pax> VAR=xyzzy
pax> if [[ -e $VAR ]] ; then echo yes ; fi
pax> VAR=/tmp
pax> if [[ -e $VAR ]] ; then echo yes ; fi
yes
In other words, just use:
换句话说，只需使用：
if [[ -e "$var" ]] ; then
echo "'$var' exists"
fi
回答by Slava Semushin
if [ -n "$var" -a -e "$var" ]; then
do something ...
fi
回答by user123444555621
Simply quote your variable:
简单地引用你的变量：
[ -e "$VAR" ]
This evaluates to [ -e "" ]if $VARis empty.
这评估为[ -e "" ]if$VAR为空。
Your version does not work because it evaluates to [ -e ]. Now in this case, bash simply checks if the single argument (-e) is a non-empty string.
您的版本不起作用，因为它评估为[ -e ]. 现在在这种情况下，bash 只是检查单个参数 ( -e) 是否为非空字符串。
From the manpage:
从联机帮助页：
test and [ evaluate conditional expressions using a set of rules based on the number of arguments. ...
1 argument
The expression is true if and only if the argument is not null.
test 和 [ 使用一组基于参数数量的规则来评估条件表达式。...
1 个参数
当且仅当参数不为空时，表达式为真。
(Also, this solution has the additional benefit of working with filenames containing spaces)
（此外，此解决方案还具有使用包含空格的文件名的额外好处）
回答by rororo
I found an answer now. Thanks for your suggestions!
我现在找到了答案。感谢您的建议！
for e in ./*.cutoff.txt; do
if grep -q -E 'COX1|Cu-oxidase' $e
then
echo xyz >$e.match.txt
else
echo
fi
if grep -q -E 'AMO' $e
then
echo abc >$e.match.txt
else
echo
fi; done
Any comments on that? It seems inefficient to grep twice, but it works...
对此有何评论？grep 两次似乎效率低下，但它有效......
