bash从函数设置多个变量的方式

本文介绍了bash从函数设置多个变量的方式的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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 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 ^ ^ 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 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/kshfunction foo { echo "this:that:the:other"}IFS=":"foo | read one two three fourecho "one = $one"echo "two = $two"echo "three = $three"echo "four = $four"This produces:one = thistwo = thatthree = thefour = otherWorks like a charm. Let's try the same program using BASH instead of Kornshell:#!/bin/bashfunction foo { echo "this:that:the:other"}IFS=":"foo | read one two three fourecho "one = $one"echo "two = $two"echo "three = $three"echo "four = $four"This produces:one = two = three = four = The pipe to the read doesn't work at all. Let's try this as a hereis read:#!/bin/bashfunction foo { echo "this:that:the:other"}IFS=":"read one two three four<<<$(foo)echo "one = $one"echo "two = $two"echo "three = $three"echo "four = $four"I get:one = this that the othertwo =three =four =That allowed $one to be set, but not the other variables. Strangely, the colons were removed from the string. Let's remove the colons from the return value:#!/bin/bashfunction foo { echo "this that the other"}read one two three four<<<$(foo)echo "one = $one"echo "two = $two"echo "three = $three"echo "four = $four"one = thistwo = thatthree = thefour = otherThat does work. Each variable is read in from function foo.What am I doing wrong? Why doesn't setting IFS seem to work the way I would think it would in BASH? Or is there a better way to do this? 解决方案 You have to quote:#!/bin/bashfunction foo { echo "this:that:the:other"}IFS=":"read one two three four<<<"$(foo)" ^ ^When executed it returns:$ ./aone = thistwo = thatthree = thefour = otherRegarding this one not working:#!/bin/bash... IFS=":"foo | read one two three fourecho "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 = thistwo = thatthree = thefour = other 这篇关于bash从函数设置多个变量的方式的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！