存储在变量中的赋值语句导致在Bash中找不到命令错误 [英] Assignment statement stored in a variable causes command not found error in Bash

问题描述

例如:

#!/bin/bash
sss='ls -l'
$sss
ttt='a=100'
$ttt

ls 的输出正确，但是，赋值语句将输出错误消息:

The output of ls is correct, however, the assignment statement will output an error message:

第5行:a = 100:未找到命令

line 5: a=100: command not found

为什么有区别?

如果分配不是命令，那是什么?我的意思是显式的 a = 100 和从变量扩展的 a = 100 有什么区别，我的意思是，bash看到的是相同的 a = 100 ，对不对?为什么他们会有不同的解释?

If assignment is not command, what is it? I mean what is the difference between explicit a=100 and a=100 expanded from variable, I mean, the bash sees the same thing a=100, right? Why they got different interpretation?

推荐答案

这是因为变量扩展的输出是作为命令运行的，因此就像在字面上直接插入了内容一样，在命令行中被替换了.

That's because the output from variable expansion is run as a command, precisely get replaced in the command line as if you have inserted the content literally.

在这里，您有 ttt ='a = 100'，因此当您接下来执行 $ ttt 时，它会简单地扩展为 a = 100 ，因为这是要运行的命令，它是唯一存在的参数.该错误是由于明显的事实，即这不是有效的命令.

Here, you have ttt='a=100', so when you do $ttt next, it will be simple expanded as a=100 and as this would be the command to run being the only parameter present. And the error is due to the obvious fact that this is not a valid command.

您可以想象，可以使用一些实际有效的命令来扩展该扩展，以将该扩展作为该命令的参数(例如 echo $ ttt ).

You can tack the expansion with some actual-valid command to get the expansion as that command's argument (e.g. echo $ttt), as you can imagine.

如果您需要进行这样的分配，请利用 declare :

If you ever need to do assignments like that, leverage declare:

$ ttt='a=100'

$ declare "$ttt"

$ echo "$a"
100

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