如何使用bash将函数的输出分配给变量？

I have a bash function that produces some output:

我有一个bash函数，可以产生一些输出：

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")

即。 VAR =扫描（当然这不起作用 - 它使VAR等于字符串“scan”）

3 个解决方案

#1

VAR=$(scan)

Exactly the same way as for programs.

与程序完全相同。

#2

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。这些功能也可用于子壳和传递，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，是实现所需结果的直接方式。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（通常是有用的）副作用是它将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删除。

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'.

如果函数试图实现的是将字符串“返回”变量，使用bash v4.3及更高版本，可以使用所谓的nameref。 Namerefs允许函数获取一个或多个变量输出参数的名称。您可以将事物分配给nameref变量，就好像您更改了它指向/ references的变量一样。

function scan3() {
    local -n outvar=$1    # -n makes it a nameref.
    local nl=$'\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:

打印：

output

===

This form has a few advantages. Namely, it allows your function to modify the environment of the caller without using global variables everywhere.

这种形式有一些优点。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：如果你的函数很大程度上依赖于bash内置函数，使用namerefs可以极大地提高程序的性能，因为它可以避免创建一个刚刚抛弃的子shell。这通常对于经常重复使用的小功能更有意义，例如，函数以echo“$ returnstring”结尾

This is relevant. https://*.com/a/38997681/5556676

这是相关的。 https://*.com/a/38997681/5556676

#3

I think init_js should use declare instead of local!

我认为init_js应该使用declare而不是local！

function scan3() {
    declare -n outvar=$1    # -n makes it a nameref.
    local nl=$'\x0a'
    outvar="output${nl}${nl}"  # two total. quotes preserve newlines
}

#1

VAR=$(scan)