bash 如何在bash中回显包含未转义美元符号的变量
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How to echo a variable containing an unescaped dollar sign in bash
提问by Not22
If I have a variable containing an unescaped dollar sign, is there any way I can echo the entire contents of the variable?
如果我有一个包含未转义美元符号的变量，有什么办法可以回显变量的全部内容吗？
For example something calls a script:
例如某事调用脚本：
./script.sh "test1$test2"
./script.sh "test1$test2"
and then if I want to use the parameter it gets "truncated" like so:
然后，如果我想使用该参数，它会像这样被“截断”：
echo ${1}
test1
echo ${1}
test1
Of course single-quoting the varaible name doesn't help. I can't figure out how to quote it so that I can at least escape the dollar sign myself once the script recieves the parameter.
当然，单引号变量名无济于事。我不知道如何引用它，以便我至少可以在脚本收到参数后自己转义美元符号。
采纳答案by Ignacio Vazquez-Abrams
The variable is replaced before the script is run.
在脚本运行之前替换变量。
./script.sh 'test1$test2'
回答by
The problem is that script receives "test1" in the first place and it cannot possibly know that there was a reference to an empty (undeclared) variable. You have to escape the $before passing it to the script, like this:
问题是脚本首先接收“test1”，它不可能知道存在对空（未声明）变量的引用。您必须$在将其传递给脚本之前转义，如下所示：
./script.sh "test1$test2"
Or use single quotes 'like this:
或者'像这样使用单引号：
./script.sh 'test1$test2'
In which case bash will not expand variables from that parameter string.
在这种情况下，bash 不会从该参数字符串中扩展变量。
回答by ghostdog74
by using single quotes , meta characters like $will retain its literal value. If double quotes are used, variable names will get interpolated.
通过使用单引号，元字符 like$将保留其文字值。如果使用双引号，变量名将被插入。
回答by Benoit
As Ignacio told you, the variable is replaced, so your scripts gets ./script.sh test1as values for $0and $1.
正如 Ignacio 告诉您的那样，该变量已被替换，因此您的脚本将./script.sh test1作为$0和 的值$1。
But even in the case you had used literal quotes to pass the argument, you shoudl always quote "$1"in your echo "${1}". This is a good practice.
但即使在您使用文字引号传递参数的情况下，您也应该始终"$1"在echo "${1}". 这是一个很好的做法。
