awk命令拆分第n个字段 [英] awk command to split nth field

问题描述

我正在学习AWK，并且正在尝试一些有关内置字符串函数的练习.

I am learning AWK and was trying some exercises on built-in string functions.

这是我的锻炼:

我有一个包含以下内容的文件

I have a file containing as below

RecordType:83    
1,2,3,a|x|y|z,4,5

我想要的输出如下:

RecordType:83       
1,2,3,a,4,5         
1,0,0,x,4,5         
1,0,0,y,4,5       
1,0,0,z,4,5

我为上面的输出写了一个awk命令.

I wrote an awk command for the above output.

awk -F',' '$1 ~ /RecordType:83/{print $0}

$1 == 1{

split($4,splt,"|")

for(i in splt)

{ 

if(i==1) 

print $1,$2,$3,splt[i],$5,$6 

else 

print $1,0,0,splt[i],$5,$6

} 

}' OFS=, file_name

上面的命令看起来很笨拙.有什么办法可以最小化命令?

The above command looks so clumsy. Is there any way minimizing the command?

预先感谢

推荐答案

我能管理的最短的一线飞机:

The shortest possible one-liner I could manage:

awk -F, 'NR>1{n=split($4,a,"|");for(;i++<n;){$4=a[i];print;$2=$3=0}}NR==1' OFS=, file
RecordType:83    
1,2,3,a,4,5
1,0,0,x,4,5
1,0,0,y,4,5
1,0,0,z,4,5

更具可读性的脚本(推荐):

BEGIN {
    FS=OFS=","                                       # Comma delimiter 
}
NR==1 {                                              # If the first line in file
    print $0                                         # Print the whole line
    next                                             # Skip to next line
}
{
    n=split($4,a,"|")                                # Split field four on |
    for(i=1;i<=n;i++)                                # For each sub-field
        print $1,i==1?$2OFS$3:"0"OFS"0",a[i],$5,$6   # Print the output
}

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