bash - Bash 使用第一行计数输出将空格添加到以下行

Question

我这里有这段代码

printf '$request1 = "select * from whatever where this = that and active = 1 order by something asc";\n' 
| perl -pe 's/select/SELECT/gi ; s/from/\n   FROM/gi ; s/where/\n      WHERE/gi ; s/and/\n      AND/gi ; s/order by/\n         ORDER BY/gi ; s/asc/ASC/gi ; s/desc/DESC/gi ;' 
| awk '{gsub(/\r/,"");printf "%s\n%d",$0,length($0)}'

它目前产生这样的输出

$request1 = "SELECT * 
22   FROM whatever 
17      WHERE this = that 
24      AND active = 1 
21         ORDER BY something ASC";

我想计算第一行（22）的数量并将该数量的空白添加到每一行。

Answer 1

假设您不想打印数字，请将 AWK 命令更改为：

awk 'NR == 1 {pad = length($0); print} NR > 1 {gsub(/\r/,""); printf "%*s%s\n", pad, " ", $0}'

输出：

$request1 = "SELECT * 
                         FROM whatever 
                            WHERE this = that 
                            AND active = 1 
                               ORDER BY something ASC";