Linux 内联 if shell 脚本
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Inline if shell script
提问by London
Is it possible to execute shell script in command line like this :
是否可以像这样在命令行中执行 shell 脚本：
counter=`ps -ef | grep -c "myApplication"`; if [ $counter -eq 1 ] then; echo "true";
>
Above example is not working I get only >character not the result I'm trying to get, that is "true"
上面的例子不起作用我只得到>字符而不是我想要得到的结果，那就是“真实”
When I execute ps -ef | grep -c "myApplicationI get 1 output. Is it possible to create result from single line in a script ? thank you
当我执行时，ps -ef | grep -c "myApplication我得到 1 个输出。是否可以从脚本中的单行创建结果？谢谢你
采纳答案by Shawn Chin
It doesn't work because you missed out fito end your ifstatement.
它不起作用，因为你错过了fi结束你的if陈述。
counter=`ps -ef | grep -c "myApplication"`; if [ $counter -eq 1 ]; then echo "true"; fi
You can shorten it further using:
您可以使用以下方法进一步缩短它：
if [ $(ps -ef | grep -c "myApplication") -eq 1 ]; then echo "true"; fi
Also, do take note the issue of ps -ef | grep ...matching itself as mentioned in @DigitalRoss' answer.
另外，请注意@DigitalRoss' answer 中ps -ef | grep ...提到的匹配自身的问题。
update
更新
In fact, you can do one better by using pgrep:
事实上，您可以使用pgrep以下方法做得更好：
if [ $(pgrep -c "myApplication") -eq 1 ]; then echo "true"; fi
回答by DigitalRoss
Yes, with syntax issues fixed
是的，已修复语法问题
That almostworked. The correct syntax is:
那几乎奏效了。正确的语法是：
counter=`ps -ef | grep -c "myApplication"`; if [ $counter -eq 1 ]; then echo "true"; fi
But note that in an expression of this sort involving psand grep, the grepwill usually match itself because the characters "grep -c Myapplication" show up in the ps listing. There are several ways around that, one of them is to grep for something like [M]yapplication.
但请注意，在此类涉及psand的表达式中grep， thegrep通常会匹配自身，因为字符“grep -c Myapplication”出现在 ps 列表中。有几种方法可以解决这个问题，其中之一是 grep 类似的东西[M]yapplication。
回答by William Pursell
Other responses have addressed your syntax error, but I would strongly suggest you change the line to:
其他回复已解决您的语法错误，但我强烈建议您将该行更改为：
test $(ps -ef | grep -c myApplication) -eq 1 && echo true
If you are not trying to limit the number of occurrences to exactly 1 (eg, if you are merely trying to check for the output line myApplication and you expect it never to appear more than once) then just do:
如果您不想将出现次数限制为恰好 1（例如，如果您只是想检查输出行 myApplication 并且您希望它永远不会出现超过一次），那么只需执行以下操作：
ps -ef | grep myApplication > /dev/null && echo true
(If you need the variable counter set for later processing, neither of these solutions will be appropriate.)
（如果您需要为以后的处理设置可变计数器，则这些解决方案都不合适。）
Using short circuited && and || operators is often much clearer than embedding if/then constructs, especially in one-liners.
使用短路 && 和 || 运算符通常比嵌入 if/then 结构更清晰，尤其是在单行中。
回答by Tagar
I was struggling to combine both multiple lines feed into command and getting its results into a variable (not a file) and come up with this solution:
我正在努力将多行输入合并到命令中并将其结果放入变量（而不是文件）中并提出以下解决方案：
FRA_PARAM="'db_recovery_file_dest'"
FRA=$(
sqlplus -S "/as sysdba" <<EOF
set echo off head off feed off newpage none pages 1000 lines 1000
select value from v$parameter where name=$FRA_PARAM;
exit;
EOF
)
Please note that single-quotes word was substituted, because otherwise I was receiving its autosubstitution to double quotes... ksh, HP-UX.
请注意单引号单词已被替换，否则我将收到它对双引号的自动替换... ksh，HP-UX。
Hopefully this will be helpful for somebody else too.
希望这对其他人也有帮助。
回答by minhas23
I am using Mac OS and following worked very well
我正在使用 Mac OS 并且以下效果很好
$ counter=`ps -ef | grep -c "myApplication"`; if [ $counter -eq 1 ]; then echo "true";fi;
true
真的
Space is needed after [ and before ]
[ ] 之后和之前需要空格
