使用bash以类似字典的格式进行迭代

本文介绍了使用bash以类似字典的格式进行迭代的处理方法，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

我在输入文件中有一个类似python的字典:

I have a python like dictionary in an input file:

$ cat test.txt db={1:['a','b','c','d'], 2:['aa','bb','cc','dd']}

字典中的每个列表只有4个索引，且该索引不小于或大于.我需要这样的结果:

Each list in dictionary only has 4 indexes not less or more. I need a result like:

one1="a" two1="b" three1="c" four1="d" one2="aa" two2="bb" three2="cc" four2="dd"

如果我们在这里使用python，我知道这很简单，但是我应该在bash脚本中完成这项工作.是否可以?如何使用bash脚本完成这项工作?

I know this is simple if we use python here, but I should do the job in bash script. Is it possible? How can I do the job using bash script?

推荐答案

您只需要剥离所有不必要的字符并循环遍历它们即可得到结果

you just need to strip off all the unnecessary characters and loop through them to get your result

#!/bin/bash db="{1:['a','b','c','d'], 2:['aa','bb','cc','dd']}" count=1 for items in `echo $db|sed 's/{//;s/}//'` do echo one${count} = `echo $items|sed 's/^.*\[//;s/\].*$//'|cut -d ',' -f1` echo two${count} = `echo $items|sed 's/^.*\[//;s/\].*$//'|cut -d ',' -f2` echo three${count} = `echo $items|sed 's/^.*\[//;s/\].*$//'|cut -d ',' -f3` echo four${count} = `echo $items|sed 's/^.*\[//;s/\].*$//'|cut -d ',' -f4` echo '' count=`expr $count + 1` done

输出

one1 = 'a' two1 = 'b' three1 = 'c' four1 = 'd' one2 = 'aa' two2 = 'bb' three2 = 'cc' four2 = 'dd'