Bash 使用 subshell 和 substring 进行错误替换
一个人为的例子......考虑到
FOO="/foo/bar/baz"
这个工作（在bash中）
BAR=$(basename $FOO) # result is BAR="baz"
BAZ=${BAR:0:1} # result is BAZ="b"
这不是
BAZ=${$(basename $FOO):0:1} # result is bad substitution
我的问题是哪个规则导致这个[子shell替换]评估不正确？在 1 跳中执行此操作的正确方法（如果有）是什么？
A contrived example... given
FOO="/foo/bar/baz"
this works (in bash)
BAR=$(basename $FOO) # result is BAR="baz"
BAZ=${BAR:0:1} # result is BAZ="b"
this doesn't
BAZ=${$(basename $FOO):0:1} # result is bad substitution
My question is which rule causes this [subshell substitution] to evaluate incorrectly? And what is the correct way, if any, to do this in 1 hop?
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评论（6）
首先，请注意，当您这样说时：
BAZ构造中的第一位是BAR，而不是您想要采用的值的第一个字符。因此，即使 bash 允许变量名称包含任意字符，第二个表达式中的结果也不会是您想要的。然而，至于阻止这种情况的规则，请允许我引用 bash 手册页：
然后稍后：
然后当它定义您要询问的语法时：
因此，手册页中阐明的规则说 < code>${foo:x:y} 构造必须有一个参数作为第一部分，并且参数只能是名称、数字或少数特殊参数字符之一。
$(basename $FOO)不是参数允许的可能性之一。至于在一项作业中执行此操作的方法，请使用其他响应中提到的其他命令的管道。
First off, note that when you say this:
the first bit in the construct for
BAZisBARand not the value that you want to take the first character of. So even if bash allowed variable names to contain arbitrary characters your result in the second expression wouldn't be what you want.However, as to the rule that's preventing this, allow me to quote from the bash man page:
Then a bit later:
And later when it defines the syntax you're asking about:
So the rules as articulated in the manpage say that the
${foo:x:y}construct must have a parameter as the first part, and that a parameter can only be a name, a number, or one of the few special parameter characters.$(basename $FOO)is not one of the allowed possibilities for a parameter.As for a way to do this in one assignment, use a pipe to other commands as mentioned in other responses.
参数替换的修改形式，例如
${parameter#word}只能修改参数，不能修改任意单词。在这种情况下，您可以将
basename的输出通过管道传输到 dd 命令，例如（如果您想要更高的计数，请增加
count而不是bs，否则您获得的字节数可能少于请求的字节数。）在一般情况下，无法在一次作业中执行此类操作。
Modified forms of parameter substitution such as
${parameter#word}can only modify a parameter, not an arbitrary word.In this case, you might pipe the output of
basenameto a dd command, like(If you want a higher count, increase
countand notbs, otherwise you may get fewer bytes than requested.)In the general case, there is no way to do things like this in one assignment.
它失败是因为
${BAR:0:1}是变量扩展。 Bash 希望在${之后看到变量名，而不是值。我不知道如何用单个表达式来做到这一点。
It fails because
${BAR:0:1}is a variable expansion. Bash expects to see a variable name after${, not a value.I'm not aware of a way to do it in a single expression.
正如其他人所说， ${} 的第一个参数需要是变量名。但是您可以使用另一个子 shell 来近似您想要执行的操作。
代替：
使用：
As others have said, the first parameter of ${} needs to be a variable name. But you can use another subshell to approximate what you're trying to do.
Instead of:
Use:
针对您的人为示例的人为解决方案：
如
A contrived solution for your contrived example:
as in
${string:0:1},string 必须是变量名
例如：
FOO="/foo/bar/baz"
baz="foo"
BAZ=
eval echo '${'"$(basename $FOO )"':0:1}'echo $BAZ
结果是 'f'
${string:0:1},string must be a variable name
for example:
FOO="/foo/bar/baz"
baz="foo"
BAZ=
eval echo '${'"$(basename $FOO)"':0:1}'echo $BAZ
the result is 'f'