bash 如果命令输出包含某个字符串，我如何测试（在一行中）？
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How do I test (in one line) if command output contains a certain string?
提问by Dane O'Connor
In one line of bash, how do I return an exit status of 0 when the output of /usr/local/bin/monit --versiondoesn't contain exactly 5.5and an exit status of 1 when it does?
在一行 bash 中，当输出/usr/local/bin/monit --version不完全包含时如何返回退出状态为 0，而在包含时返回5.5退出状态为 1？
回答by ruakh
! /usr/local/bin/monit --version | grep -q 5.5
(grepreturns an exit-status of 0 if it finds a match, and 1 otherwise. The -qoption, "quiet", tells it not to print any match it finds; in other words, it tells grepthat the only thing you want is its return-value. The !at the beginning inverts the exit-status of the whole pipeline.)
（grep如果找到匹配，则返回退出状态 0，否则返回 1。-q选项“quiet”告诉它不打印它找到的任何匹配；换句话说，它告诉grep你唯一想要的是它的返回-value.!开头反转整个管道的退出状态。）
Edited to add:Alternatively, if you want to do this in "pure Bash" (rather than calling grep), you can write:
编辑添加：或者，如果您想在“纯 Bash”（而不是调用grep）中执行此操作，您可以编写：
[[ $(/usr/local/bin/monit --version) != *5.5* ]]
([[...]]is explained in §3.2.4.2 "Conditional Constructs" of the Bash Reference Manual. *5.5*is just like in fileglobs: zero or more characters, plus 5.5, plus zero or more characters.)
（在Bash 参考手册的第 3.2.4.2 节“条件构造”中进行了[[...]]解释。就像在 fileglobs 中一样：零个或多个字符，加上，加上零个或多个字符。）*5.5*5.5
回答by perreal
[ $(/usr/local/bin/monit --version) == "5.5" ]
eg-1: check for success
eg-1：检查是否成功
[ $(/usr/local/bin/monit --version) == "5.5" ] && echo "OK"
eg-2: check for failure
eg-2：检查失败
[ $(/usr/local/bin/monit --version) == "5.5" ] || echo "NOT OK"
or, to just check if the output contains 5.5:
或者，只检查输出是否包含5.5：
[[ $(/usr/local/bin/monit --version) =~ "5.5" ]] || echo "NOT OK"
回答by Matzy
Test the return value of grep:
测试grep的返回值：
sudo service xyz status | grep 'not' &> /dev/null
if [ $? == 0 ]; then
echo "whateveryouwant"
fi
I would recommend cron, it works fine with SALT stack
我会推荐 cron，它适用于 SALT 堆栈
