使用bash以类似字典的格式进行迭代 [英] Using bash for iterating in dictionary like formats

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'

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