linux - 比较两个逗号分隔的字符串并列出共同的值

Question

我比较了两个逗号分隔列表(主列表和输入列表)并列出了它们之间的共同值(结果)，同时保留了主列表中元素的顺序。例如:

案例一:

master="common,city,country"
input="city,country"

result="city,country"

案例二:

master="common,city,country"
input="country,pig,cat,common"

result="common,country"

案例三:

master="common,city,country"
input="pigs,cars,train"

result="nothing found"

这是我尝试过的:

result="$(awk -F, -v master_list=$master'{ for (i=1;i<=NF;i++) { if (master_list~ $i) { echo $i } } } END ' <<< $input)"

Answer 1

这是一个 awk-oneliner 解决方案:

awk -v RS=",|\n" 'NR==FNR{a[$0]=1;next}
    {a[$0]++}a[$0]>1{r=r?r","$0:$0}
    END{print r?r:"Nothing found"}' <(<<< $master) <(<<<$input)

测试你的 3 个案例:

案例一

kent$ master="common,city,country"
kent$ input="city,country"
kent$ result=$(awk -v RS=",|\n" 'NR==FNR{a[$0]=1;next}{a[$0]++}a[$0]>1{r=r?r","$0:$0}END{print r?r:"Nothing found"}' <(<<< $master) <(<<<$input))
kent$ echo $result
city,country

案例二

kent$ master="common,city,country"
kent$ input="country,pigs,cat,common"
kent$ result=$(awk -v RS=",|\n" 'NR==FNR{a[$0]=1;next}{a[$0]++}a[$0]>1{r=r?r","$0:$0}END{print r?r:"Nothing found"}' <(<<< $master) <(<<<$input))
kent$ echo $result
country,common

案例三

kent$ master="common,city,country"
kent$ input="pigs,cars,train"
kent$ result=$(awk -v RS=",|\n" 'NR==FNR{a[$0]=1;next}{a[$0]++}a[$0]>1{r=r?r","$0:$0}END{print r?r:"Nothing found"}' <(<<< $master) <(<<<$input))
kent$ echo $result
Nothing found