为什么这个bash代码不起作用？

Question

x="a=b"
`echo $x`
echo $a

I expect the second line to generate "a=b", and execute it in the context of the main shell, resulting in a new variable a with value b . However, what I really get (if I enter the commands manually) is the error message after the second line, bash: a=b: command not found

Why is that so?

Answer 1

Try

eval $x

(And we need 30 characters for this answer to be posted)

Answer 2

What your first echo line does is running in a subshell and returns its value to the callee.. The same result is achieved using $() and is - by the way - easier to use than backticks.

So, what you are doing is first running echo $x (which returns a=b ). And, because of the backticks, a=b is returned to the shell that tries to run that line as a command which - obviously - won't work.

Try this in a shell:

$(echo ls)

And you will clearly see what is happening.

Answer 3

It's because of the order in which bash parses the command line. It looks for variable definitions (eg a=b ) before performing variable and command substitution (eg commands in backticks). Because of this, by the time echo $x is replaced by a=b , it's too late for bash to see this as a variable definition and it's parsed as a command instead. The same thing would've happened if you'd just used $x as the command (instead of echo in backticks). As in @mvds's answer, the eval command can be used to force the command to be reparsed from the beginning, meaning that it'll be recognized as a variable definition:

$ x="a=b"
$ `echo $x`
-bash: a=b: command not found
$ $(echo $x)  # Exact same thing, but with cleaner syntax
-bash: a=b: command not found
$ $x  # This also does the same thing, but without some extra steps
-bash: a=b: command not found
$ eval "$x"  # This will actually work
$ echo $a
b
$ a=  # Start over
$ eval "$(echo "$x")"  # Another way of doing the same thing, with extra steps
$ echo $a
b

Note that when using eval I've put all of the references to $x in double-quotes -- this is to prevent the later phases of bash parsing (eg word splitting) from happening twice , since bash will finish its regular parsing process, then recognize the eval command, and then redo the entire parsing process again . It's really easy to get unexpected results from using eval , and this removes at least some of the potential for trouble.

Answer 4

Did you try $x in that funny apostrophes? Without echo , echo seems to be only for displaying string, not execute commands.