设置变量时找不到bash命令 [英] bash command not found when setting a variable

问题描述

我正在编写一个shell脚本，其中设置了几个变量，这些变量的值是命令的输出.

I am writing a shell script where I am setting few variables, whose value is the output of commands.

我得到的错误是:

$  $tag_name="proddeploy-$(date +"%Y%m%d_%H%M")"
-bash: =proddeploy-20141003_0500: command not found

现在，我确实阅读了其他类似问题，并基于它，我尝试了各种方法:

now, I did read other similar questions and based on it, I tried various things:

$ $deploy_date=date +"%Y%m%d_%H%M"
bash: =date: command not found
$ $tag_name="proddeploy-$deploy_date"
bash: proddeploy- command not found

尝试使用反引号

$ $tag_name=`proddeploy-$(date +"%Y%m%d_%H%M")`
bash: proddeploy-20141003_1734: command not found
bash: =: command not found

尝试使用$()

$ $tag_name=$(proddeploy-$(date +"%Y%m%d_%H%M"))
bash: proddeploy-20141003_1735: command not found
bash: =: command not found

但是在每种情况下，命令输出都会执行.如何使其停止执行命令输出并仅将其存储为变量?我需要它来处理ZSH和BASH.

But in every case the command output is getting executed. how do I make it to stop executing command output and just store as a variable? I need this to work on ZSH and BASH.

推荐答案

您可以使用var=string或var=$(command)定义变量.

You define variables with var=string or var=$(command).

因此，您必须删除前导$和=周围的所有其他符号:

So you have to remove the leading $ and any other signs around =:

tag_name="proddeploy-$(date +"%Y%m%d_%H%M")"

deploy_date=$(date +"%Y%m%d_%H%M")
            ^^                   ^

来自命令替换:

第二种形式`COMMAND`对于Bash或多或少已经过时，因为它 嵌套时遇到了一些麻烦(内部"反引号需要转义) 和转义字符.使用$(COMMAND)，它也是POSIX！

The second form `COMMAND` is more or less obsolete for Bash, since it has some trouble with nesting ("inner" backticks need to be escaped) and escaping characters. Use $(COMMAND), it's also POSIX!

此外，$()允许您嵌套，这可能很方便.

Also, $() allows you to nest, which may be handy.

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