使用printf表格式化的输出

Question

我尝试使用printf创建格式化输出。但我无法按照我的意愿对齐输出。我的代码是象下面这样：

#!/bin/bash 
header="%-10s %8s %13s %15s %3s\n" 
format="%-10s %8s %13s %15s %3s\n" 

lvalue1="10000000000" 
rvalue1="10000000000" 

lvalue2="20000000000" 
rvalue2="20000000000" 

lvalue3="10000" 
rvalue3="10000" 

lvalue4="300000000" 
rvlaue4="300000000" 

lvalue5="2000" 
rvalue5="2000" 

value6="1500" 

value7="2200" 

lvalue8="2300" 
rvalue8="2300" 


formula1="formula1" 
formula2="formula2" 
formula3="formula3" 
formula4="formula4" 
formula5="formula5" 
formula6="formula6" 
formula7="formula7" 
formula8="formula8" 

check1="Total Received Packets" 
check2="Total Transmitted Packets" 
check3="Packets Lost" 
check4="Media Access Control Address" 
check5="IPv4 Default Gateway" 
check6="Input Rate (bits/sec)" 
check7="Output Rate (bits/sec)" 
check8="Ignored Packets" 


printf "$header" "CHECK" "FORMULA" "LVALUE" "RVALUE" "PASSED" 
printf "$format" \ 
"$check1" "$formula1" "$lvalue1" "$rvalue1" "y/n" \ 
"$check2" "$formula2" "$lvalue2" "$rvalue2" "y/n" \ 
"$check3" "$formula3" "$lvalue3" "$rvalue3" "y/n" \ 
"$check4" "$formula4" "$lvalue4" "$rvalue4" "y/n" \ 
"$check5" "$formula5" "$lvalue5" "$rvalue5" "y/n" \ 
"$check6" "$formula6" "$value6" "    " "y/n" \ 
"$check7" "$formula7" "$value7" "    " "y/n" \ 
"$check8" "$formula8" "$lvalue8" "$rvalue8" "y/n" 

和输出的样子：

CHECK  FORMULA  LVALUE   RVALUE PASSED 
Total Received Packets formula1 10000000000  10000000000 y/n 
Total Transmitted Packets formula2 20000000000  20000000000 y/n 
Packets Lost formula3   10000   10000 y/n 
Media Access Control Address formula4  300000000     y/n 
IPv4 Default Gateway formula5   2000   2000 y/n 
Input Rate (bits/sec) formula6   1500     y/n 
Output Rate (bits/sec) formula7   2200     y/n 
Ignored Packets formula8   2300   2300 y/n 

正如你看到的，字段不正确对齐。那么我应该在哪里修改以更好地查找此表？

由于

Answer 1

printf字段长度是最小长度。你第一个领域太短，所以，例如：

format="%-30s %8s %13s %15s %3s\n"