如何使用 bash 将函数的输出分配给变量？
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How can I assign the output of a function to a variable using bash?
提问by Brent
I have a bash function that produces some output:
我有一个生成一些输出的 bash 函数：
function scan {
echo "output"
}
How can I assign this output to a variable?
如何将此输出分配给变量？
ie. VAR=scan (of course this doesn't work - it makes VAR equal the string "scan")
IE。VAR=scan（当然这不起作用 - 它使 VAR 等于字符串“scan”）
回答by Robert Obryk
VAR=$(scan)
Exactly the same way as for programs.
与程序完全相同的方式。
回答by init_js
You may use bash functions in commands/pipelines as you would otherwise use regular programs. The functions are also available to subshells and transitively, Command Substitution:
您可以在命令/管道中使用 bash 函数，就像使用常规程序一样。这些函数也可用于 subshell 和可传递的命令替换：
VAR=$(scan)
Is the straighforward way to achieve the result you want in most cases. I will outline special cases below.
在大多数情况下，这是实现您想要的结果的直接方式。我将在下面概述特殊情况。
Preserving trailing Newlines:
保留尾随换行符：
One of the (usually helpful) side effects of Command Substitution is that it will strip any number of trailing newlines. If one wishes to preserve trailing newlines, one can append a dummy character to output of the subshell, and subsequently strip it with parameter expansion.
命令替换的一个（通常有用的）副作用是它会删除任意数量的尾随换行符。如果希望保留尾随换行符，可以在子 shell 的输出中附加一个虚拟字符，然后通过参数扩展将其剥离。
function scan2 () {
local nl=$'\x0a'; # that's just \n
echo "output${nl}${nl}" # 2 in the string + 1 by echo
}
# append a character to the total output.
# and strip it with %% parameter expansion.
VAR=$(scan2; echo "x"); VAR="${VAR%%x}"
echo "${VAR}---"
prints (3 newlines kept):
打印（保留 3 个换行符）：
output
---
Use an output parameter: avoiding the subshell (and preserving newlines)
使用输出参数：避免使用子shell（并保留换行符）
If what the function tries to achieve is to "return" a string into a variable , with bash v4.3 and up, one can use what's called a nameref. Namerefs allows a function to take the name of one or more variables output parameters. You can assign things to a nameref variable, and it is as if you changed the variable it 'points to/references'.
如果该函数试图实现的是将一个字符串“返回”到一个变量中，在 bash v4.3 及更高版本中，可以使用所谓的 a nameref。Namerefs 允许函数采用一个或多个变量的名称作为输出参数。您可以将事物分配给 nameref 变量，就好像您更改了它“指向/引用”的变量一样。
function scan3() {
local -n outvar= # -n makes it a nameref.
local nl=$'\x0a'
outvar="output${nl}${nl}" # two total. quotes preserve newlines
}
VAR="some prior value which will get overwritten"
# you pass the name of the variable. VAR will be modified.
scan3 VAR
# newlines are also preserved.
echo "${VAR}==="
prints:
印刷：
output
===
This form has a few advantages. Namely, it allows your function to modify the environment of the caller without using global variables everywhere.
这种形式有几个优点。也就是说，它允许您的函数修改调用者的环境，而无需在任何地方使用全局变量。
Note: using namerefs can improve the performance of your program greatly if your functions rely heavily on bash builtins, because it avoids the creation of a subshell that is thrown away just after. This generally makes more sense for small functions reused often, e.g. functions ending in echo "$returnstring"
注意：如果您的函数严重依赖 bash 内置函数，则使用 namerefs 可以极大地提高程序的性能，因为它避免了创建之后被丢弃的子shell。这通常对于经常重用的小函数更有意义，例如以echo "$returnstring"
This is relevant. https://stackoverflow.com/a/38997681/5556676
回答by Hamid Reza Hasani
I think init_js should use declare instead of local!
我认为 init_js 应该使用声明而不是本地！
function scan3() {
declare -n outvar= # -n makes it a nameref.
local nl=$'\x0a'
outvar="output${nl}${nl}" # two total. quotes preserve newlines
}
