如何在bash shell中重定向命令输出? [英] how to redirect command output in bash shell?

问题描述

我在编写一个小的bash命令时遇到了问题.基本上，我想回显wrapper命令并将真实命令的输出重定向到日志文件.

I'm having a problem writing a small bash command. Basically I want to echo the wrapper command and redirect the output of the real command to a log file.

.bashrc中的类似内容不起作用-输出仍进入控制台.

Something like this in my .bashrc doesn't work -- the output still gets to the console.

cmd="some_command >& output.log";
echo $cmd;
$cmd;

但是下面的工作-输出直接进入日志文件.

But the following works -- the output is directed into the log file.

cmd = "some_command";
echo $cmd" >& output.log";
$cmd >& output.log;

第一种方法有什么问题?如何解决?

What is wrong with the first method? How to fix it?

谢谢！

推荐答案

第一种方法有什么问题?

What is wrong with the first method?

在变量中包含重定向运算符时，shell不会将其视为特殊字符.而是将它们视为有关程序的参数.

When you include the redirection operators within a variable, the shell doesn't treat those as special. Instead those are considered as arguments to the program in question.

一种解决方案是利用eval:

cmd="some command >& output.log";
eval $cmd;

顺便说一句，以下是错误的:

As an aside, the following is wrong:

cmd = "some command";

在变量分配中，=周围不能有空格.

You cannot have spaces around = in a variable assignment.

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