I'm trying to get stored in a variable the output of a command chain with awk, but the result is incomplete,
我试图用awk将命令链的输出存储在变量中，但结果不完整，
this is what im trying to do.
这就是我想要做的事情。
nameserver="nas_server -list all | awk '{print $6}'"
eval $nameserver
nameserverreal=$nameserver |awk '/encap|nameserver_/{ print }'
eval $nameserverreal
I'm using this command nas_server and with awk print 6, only get some output from my command nas_server.
我正在使用这个命令nas_server和awk print 6，只从我的命令nas_server获取一些输出。
What i need is to filter later the output only with "nameserver_" and storage the output in a variable, so i can print it, and use it later convined with other commands.
我需要的是稍后使用“nameserver_”过滤输出并将输出存储在变量中，这样我就可以打印它，并在以后用其他命令进行后续处理。
1 个解决方案
#1
2
You appear to be confusing storing the output of commands with storing the text of the commands themselves (which is almost always a bad idea). I'm not sure exactly what you're trying to do (or what the output of nas_server -list all looks like), but I suspect you want something like this:
你似乎混淆了存储命令输出和存储命令本身的文本（这几乎总是一个坏主意）。我不确定你要做什么（或者nas_server -list的输出都是什么样子），但我怀疑你想要这样的东西：
nameserver="$(nas_server -list all | awk '{print $6}')" # $() captures the output of a command
echo "$nameserver" # Double-quote all variable references to avoid parsing weirdness!
nameserverreal="$(echo "$nameserver" |awk '/encap|nameserver_/{ print }')"
echo "$nameserverreal"
Here's a simplified version:
这是一个简化版本：
nameserverreal="$(nas_server -list all | awk '$6 ~ /encap|nameserver_/ {print $6}'"
Oh, and anytime you're tempted to use eval in a shell script, it's a sign that something has gone horribly wrong.
哦，任何时候你都想在shell脚本中使用eval，这表明事情已经发生了可怕的错误。
#1
2
You appear to be confusing storing the output of commands with storing the text of the commands themselves (which is almost always a bad idea). I'm not sure exactly what you're trying to do (or what the output of nas_server -list all looks like), but I suspect you want something like this:
你似乎混淆了存储命令输出和存储命令本身的文本（这几乎总是一个坏主意）。我不确定你要做什么（或者nas_server -list的输出都是什么样子），但我怀疑你想要这样的东西：
nameserver="$(nas_server -list all | awk '{print $6}')" # $() captures the output of a command
echo "$nameserver" # Double-quote all variable references to avoid parsing weirdness!
nameserverreal="$(echo "$nameserver" |awk '/encap|nameserver_/{ print }')"
echo "$nameserverreal"
Here's a simplified version:
这是一个简化版本：
nameserverreal="$(nas_server -list all | awk '$6 ~ /encap|nameserver_/ {print $6}'"
Oh, and anytime you're tempted to use eval in a shell script, it's a sign that something has gone horribly wrong.
哦，任何时候你都想在shell脚本中使用eval，这表明事情已经发生了可怕的错误。