Linux shell 脚本的详细输出
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/956213/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Verbose output of shell script
提问by
I have a very simple shell script that looks as follows:
我有一个非常简单的 shell 脚本，如下所示：
clear
for i in -20 -19 -18 -17 -16 -15 -14 ... 18 19
do
echo "Nice value is $i"
nice -n $i ./app1
done
Basically, I wanna run an application with all different priority values between -20 and 19. However, when executing this script it looks as follows:
基本上，我想运行一个具有 -20 和 19 之间所有不同优先级值的应用程序。但是，在执行此脚本时，它看起来如下：
Nice value is -20
15916233
Nice value is -19
5782142
....
Nice value is 19
5731287
But I would like some kind of verbose output, that is also printing the command on the terminal so that it looks like this
但我想要某种详细的输出，也就是在终端上打印命令，使其看起来像这样
Nice value is -20
nice -n -20 ./app1
15916233
Nice value is -19
nice -n -19 ./app1
5782142
....
Nice value is 19
nice -n 19 ./app1
5731287
Is there a way to do that? Thank you!
有没有办法做到这一点？谢谢！
回答by Brian Agnew
You don't say what sort of shell you're running. If you're using sh/bash, try
你没有说你正在运行什么样的shell。如果您使用的是 sh/bash，请尝试
sh -x script_name
sh -x script_name
to run your script in a verbose/debug mode. This will dump out all the commands you execute, variable values etc. You don't want to do this normally since it'll provide a ton of output, but it's useful to work out what's going on.
在详细/调试模式下运行您的脚本。这将转储您执行的所有命令、变量值等。您通常不想这样做，因为它会提供大量输出，但了解正在发生的事情很有用。
As noted in the comments, you can add this flag to your #!/bin/bashinvocation inyour script.
如评论中所述，您可以将此标志添加到脚本中的#!/bin/bash调用中。
回答by seb
an easy way:
一个简单的方法：
for i in -20 -19 -18 -17 -16 -15 -14 ... 18 19
do
echo "Nice value is $i"
echo "nice -n $i ./app1"
nice -n $i ./app1
done
回答by Steve B.
let I=-20
while [ $I -lt 20 ]; do
echo "Nice value is $I"
nice -n $I ./app1
let I=$I+1
done
回答by Bert F
These will demonstrate 'eval' and 'set' to do what you want:
这些将演示 'eval' 和 'set' 来做你想做的事：
::::::::::::::
a.sh
::::::::::::::
#!/bin/sh
clear
i=-20
while [ ${i} -lt 20 ]; do
echo "Nice value is $i"
cmd="nice -n $i ./app1"
echo ${cmd}
eval ${cmd}
i=`expr ${i} + 1`
done
::::::::::::::
b.sh
::::::::::::::
#!/bin/sh
clear
i=-20
while [ ${i} -lt 20 ]; do
echo "Nice value is $i"
set -x
nice -n $i ./app1
set +x
i=`expr ${i} + 1`
done
