bash从函数设置多个变量的方式 [英] bash way of setting multiple variables from a function

问题描述

我想在更顶层的 设计中编写一个shell脚本。我用Kornshell脚本来做这件事。我定义了可能返回多组变量的各种函数（例如， getopts 函数）。然后，我可以在我的主程序中使用这些函数来设置我想要的变量。

不幸的是，BASH和Kornshell似乎在处理这个实例方面存在分歧。这是我在Kornshell中运行的一个简单脚本。我的函数 foo 返回四个值。我将在这四个值中读入我的主程序中的变量：

 ＃！/ bin / ksh 
 
函数foo {
 echothis：that：the：other
} 
 
 IFS =：
 foo |阅读一个两个三四
回声一= $一
回声二= $二
回声三= $三
回声四= $四
  

这产生：

  one = this 
 two = that 
 three = 
 four = other 
  

像魅力一样工作。让我们使用BASH而不是Kornshell来试用相同的程序：

 ＃！/ bin / bash 
 
函数foo {
 echothis：that：the：other
} 
 
 IFS =：
 foo |阅读一个两个三四
回声一= $一
回声二= $二
回声三= $三
回声四= $四
  

这产生：

  one = 
 two = 
 three = 
 four = 
  

读取的管道根本不起作用。让我们试试这个 hereis 阅读：

 ＃！/ bin / bash 
 
函数foo {
 echothis：that：other：
} 
 
 IFS =：
读取一个两个三个四个<< < $（foo）
 echoone = $ one
 echotwo = $ two
 echothree = $ three
 echofour = $ four 
  

我得到：

  one =这表示其他
 two = 
 three = 
 four = 
  

允许设置 $ 1 ，但不能设置其他变量。奇怪的是，冒号被从字符串中删除。让我们从返回值中删除冒号：

 ＃！/ bin / bash 
 
 function foo { 
 echothis that the other
} 
 
 read one two three four <<< $（foo）
 echoone = $ one
 echotwo = $ two
 echothree = $ three
 echofour = $ four
 
 one = this 
 two = that 
 three = 
 four = other 
  

这确实有效。每个变量都从函数 foo 读入。

我做错了什么？为什么不设置 IFS 似乎按照我认为在BASH中的方式工作？或者有更好的方法来做到这一点？

解决方案

您必须引用：

 ＃！/ bin / bash 
 
函数foo {
 echothis：that：the：other
} 
 
 IFS =：
读取一二三四<<<$（foo）
 ^ ^ 
   
 
 
 
执行时会返回：
 
 $ $ p $  $ ./ a 
 one = this 
 two = that 
 three = 
 four = other 
  

---

关于这个不起作用：

 ＃ ！/ bin / bash 
 
 ... 
 
 IFS =：
 foo |阅读一二三四
 echoone = $ one
  

我想这是因为 IFS 定义在与 read 不同的shell中。此外，管 foo 到 read 似乎不是将字符串赋给<$ c $在bash中读取

为此，请打开一个新shell，如 bash脚本，从标准输入管道读取值：

  $ foo | {IFS =：;读一两三四;回声一=一美元;回声two = $ 2;回声三=三美元;回声四=四美元; } 
 one = this 
 two = that 
 three = 
 four = other 
  

I want to write a shell script in a more top downish design. I've use to do this with Kornshell scripts. I define various functions that may return multiple set of variables (for example, a getopts function). I can then use these functions in my main program to set the variables I want.

Unfortunately, BASH and Kornshell seem to diverge in the way they handle this instance. Here's a simple script I run in the Kornshell. My function foo returns four values. I'll read in these four values into variables in my main program:

#!/bin/ksh

function foo {
    echo "this:that:the:other"
}

IFS=":"
foo | read one two three four
echo "one = $one"
echo "two = $two"
echo "three = $three"
echo "four = $four"

This produces:

one = this
two = that
three = the
four = other

Works like a charm. Let's try the same program using BASH instead of Kornshell:

#!/bin/bash

function foo {
    echo "this:that:the:other"
}

IFS=":"
foo | read one two three four
echo "one = $one"
echo "two = $two"
echo "three = $three"
echo "four = $four"

This produces:

one = 
two = 
three = 
four = 

The pipe to the read doesn't work at all. Let's try this as a hereis read:

#!/bin/bash

function foo {
    echo "this:that:the:other"
}

IFS=":"
read one two three four<<<$(foo)
echo "one = $one"
echo "two = $two"
echo "three = $three"
echo "four = $four"

I get:

one = this that the other
two =
three =
four =

That allowed $one to be set, but not the other variables. Strangely, the colons were removed from the string. Let's remove the colons from the return value:

#!/bin/bash

function foo {
    echo "this that the other"
}

read  one two three four<<<$(foo)
echo "one = $one"
echo "two = $two"
echo "three = $three"
echo "four = $four"

one = this
two = that
three = the
four = other

That does work. Each variable is read in from function foo.

What am I doing wrong? Why doesn't setting IFS seem to work the way I would think it would in BASH? Or is there a better way to do this?

解决方案

You have to quote:

#!/bin/bash

function foo {
    echo "this:that:the:other"
}

IFS=":"
read one two three four<<<"$(foo)"
                          ^      ^

When executed it returns:

$ ./a
one = this
two = that
three = the
four = other

---

Regarding this one not working:

#!/bin/bash

...    

IFS=":"
foo | read one two three four
echo "one = $one"

I guess it is because IFS is defined in a different shell than the one having read. Also, to pipe foo to read doesn't seem to be the way to give a string to read in bash.

To do it, open a new shell, as seen on Bash script, read values from stdin pipe:

$ foo | { IFS=":"; read one two three four; echo "one = $one"; echo "two = $two"; echo "three = $three"; echo "four = $four"; }
one = this
two = that
three = the
four = other

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