将列表转换为存储在变量中的值的数组

Question

我有一个配置文件。第一列是作业名称，其余列是与该作业关联的表。现在我正在阅读第一行。它将在 .现在我想循环这些表并创建一个 HQL 查询。我的代码的问题是表作为列表存储在变量中，当我执行 unset 命令时，它会抛出错误。任何人都可以帮助我哪里出错了。下面是配置文件、代码和所需的输出。

job1,tbl1,tbl2,tbl3
job2.tbl4,tbl5,tbl6
job3,tbl7,tbl8,tbl9

下面是我正在尝试的代码

    #!/bin/bash

while read line; do
    job_name=$(echo "$line"|awk 'BEGIN{FS=","}{print $1}')
    echo "JobName is $job_name"
    tablestring=${line#"$job_name"}
    for table in ${tablestring//,/ }; do
        echo "$table"
    done
hql="select $table as table_name, max(cycle_Date) as AS_OF_dATE FROM DB1.$table "
unset tablestring[0]
for tbl in "${tablestring[@]}"
do
hql+=$'\n'
hql+="union all select $tbl as table_name, max(cycle_Date) as AS_OF_dATE FROM DB1.$tbl"
hql+=";"
echo "$hql" >query.hql
done
done < a.configuration

以下是我希望在第一次迭代中为作业 1 以及随后为作业 2 实现的所需输出，依此类推..

  select tbl1 as table_name, max(cycle_Date) as AS_OF_dATE FROM DB1.tbl1
    union all select tbl2 as table_name, max(cycle_Date) as AS_OF_dATE FROM DB1.tbl2
    union all select tbl3 as table_name, max(cycle_Date) as AS_OF_dATE FROM DB1.tbl3;

Answer 1

所以我会用这个作为配置文件a.cfg

job[1]="jobname1 tbl1 tbl2 tbl3"
job[2]="jobname2 tbl4 tbl5 tbl6 tbl7"
job[3]="jobname3 tbl8 tbl9"

还有这个脚本

#!/bin/bash

TimeStamp=$(date +"%Y%m%d%H%M%S")

#Sourcing the configuration file. Here in the same directory as this script but you can provide the full path
. ./a.cfg

for i in ${!job[@]}; do
    #Creating the array jobtable. The job name is always index 0
    jobtable=( ${job[$i]} )
    #Using a unique name for each file
    filename=${TimeStamp}_${jobtable[0]}.hql
    echo "Processing ${jobtable[0]} using"
    for ((j=1;j<${#jobtable[@]};j++)); do
        echo "    ${jobtable[$j]}"
        #Creating the hql file
        if [[ $j -eq 1 ]]; then
            echo "select ${jobtable[$j]} as table_name, max(cycle_Date) as AS_OF_dATE FROM DB1.${jobtable[$j]} " >> $filename
        elif [[ $j -lt $((${#jobtable[@]}-1)) ]]; then
            echo " union all select ${jobtable[$j]} as table_name, max(cycle_Date) as AS_OF_dATE FROM DB1.${jobtable[$j]}" >> $filename
        else
            echo " union all select ${jobtable[$j]} as table_name, max(cycle_Date) as AS_OF_dATE FROM DB1.${jobtable[$j]};" >> $filename
        fi
    done
    #You can use the newly created hql file here using $filename. I would not delete the file though as it might be usefull to
    #troubleshoot if something is not right.
done

它确实会生成像这样的 hql 文件

select tbl1 as table_name, max(cycle_Date) as AS_OF_dATE FROM DB1.tbl1
 union all select tbl2 as table_name, max(cycle_Date) as AS_OF_dATE FROM DB1.tbl2
 union all select tbl3 as table_name, max(cycle_Date) as AS_OF_dATE FROM DB1.tbl3;

编辑：小编辑以纠正错字。