Question

我有以下文件

acc1:server1:server2:acc1234:blah:blah
acc2:server1:server5:acc4321:blah:blah
acc3:server1:server3:acc2222:blah:blah

上述文件的每一行都是数组中的元素。我只需要获得前三项accX:serverX:serverX。

如何使用ksh93在没有“：”的情况下仅拉出每个元素的前三项？

谢谢！

Answer 1

这适合你吗？

$ set -A myarr acc1:server1:server2:acc1234:blah:blah acc2:server1:server5:acc4321:blah:blah acc3:server1:server3:acc2222:blah:blah

$ $ for ele in ${myarr[@]}; do ele=${ele%:*:*:*}; echo ${ele//:/ }; done
acc1 server1 server2
acc2 server1 server5
acc3 server1 server3

Answer 2

     #!/bin/ksh

     file=~/kshinput.txt
     index=0

     while read line
     do
            myarray[$index]=${line%:*:*:*}
            ((index=index+1))
     done <"$file"


     for i in {0..$index}
     do
            echo "$i: ${myarray[$i]}"
     done

如何使用ksh93从数组中的元素中提取项目？

2 个答案: