[英]Searching within column using awk
我正在通过bashscript建立联系系统。 我在其中输入联系人的文本文件如下所示:
Sally May,may@yahoo.com,344-555-4930,Friend
Bill,Bill@yahoo.com,344-555-6543,Co-Worker
在我要求的搜索选项中(他们选择了该列之后):
echo -e"What would you like to search for:\c";;
read search
在这里,我想使用变量$ search遍历FIRST列,并以格式化的方式给我这些行。 例如:
如果他们输入(Bill),则应返回
Name Email Phone Category
Bill Bill@yahoo.com 344-555-6543 Co-Worker
如果他们输入(ll),则应返回
Name Email Phone Category
Bill Bill@yahoo.com 344-555-6543 Co-Worker
Sally May may@yahoo.com 344-555-4930 Friend
到目前为止,我一直在研究的代码行是:
awk -F, '{ if ($1 ~/$search/) print $0 }' contacts.txt | awk -F, 'BEGIN{printf "%-25s %-25s %-25s %-25s\n","Name","Email","Phone","Category"} {printf "%-25s %-25s %-25s %-25\n",$1,$2,$3,$4}' ;;
当我运行它时,它给我一个错误。 有人可以帮我解决这个问题! 我很感激
您需要使用-v
选项将变量传递给awk,并需要简化格式:
s='ll'
awk -F, -v s="$s" '$0 ~ s{$1=$1; print}' file | column -t
Sally May may@yahoo.com 344-555-4930 Friend
Bill Bill@yahoo.com 344-555-6543 Co-Worker
s='Bill'
awk -F, -v s="$s" '$0 ~ s{$1=$1; print}' file | column -t
Bill Bill@yahoo.com 344-555-6543 Co-Worker
read -p "What would you like to search for? :" patt
awk -F, 'BEGIN{printf "%-25s%-25s%-25s%-25s\n","Name","Email","Phone","Category"} $1~/'$patt'/{printf("%-25s%-25s%-25s%-25s\n",$1,$2,$3,$4)}' file
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