Question

我正在尝试迭代shell脚本中每个循环的嵌套，该脚本给出了预期的结果，但同时提供了额外的输出。

下面是代码。

 for container in ${DB_1} ${DB_2}
 do
 for container_dump in ${DB_1_bkup} ${db_2bkup}
 do
    echo "${container} and backups/${container_dump}_dump_`date +%Y-%m-%d"_"%H_%M_%S`.dump"
  done
done

我得到的输出是

DB_1 and backups/DB_1_bkup_dump_`date +%Y-%m-%d"_"%H_%M_%S`.dump
DB_1 and backups/DB_2_bkup_dump_`date +%Y-%m-%d"_"%H_%M_%S`.dump
DB_2 and backups/DB_1_bkup_dump_`date +%Y-%m-%d"_"%H_%M_%S`.dump
DB_2 and backups/DB_2_bkup_dump_`date +%Y-%m-%d"_"%H_%M_%S`.dump

我期望的是

DB_1 and backups/DB_1_bkup_dump_`date +%Y-%m-%d"_"%H_%M_%S`.dump
DB_2 and backups/DB_2_bkup_dump_`date +%Y-%m-%d"_"%H_%M_%S`.dump

有人可以帮忙吗？

Answer 1

如果我正确理解了您的问题，则不需要两个嵌套循环...
只需从bkup变量获取容器名称...

for container_dump in ${DB_1_bkup} ${db_2bkup}
do
    if [ "${container_dump}" = "${DB_1_bkup}" ]; then
        container=${DB_1}
    else
        container=${DB_2}
    fi
    echo "${container} and backups/${container_dump}_dump_`date +%Y-%m-%d"_"%H_%M_%S`.dump"
done

Shell脚本使用for循环在每次迭代中执行不同的变量

1 个答案: