Awk 如何将文件名附加到每个

我在一个文件中有32个单元格，所以我需要读取文件名并将文件名附加在每一行的开头。我该怎么做

范例

MACRO XYZ 
PIN AB 
DIRECTION INPUT ;
PIN BC
DIRECTION INPUT ;

MACRO  GEN 
PIN DECSEL_STG2[0]
DIRECTION INPUT ;
PIN DECSEL_STG1[0]
DIRECTION INPUT ;

我需要什么作为输出

MACRO XYZ 
XYZ PIN AB 
XYZ DIRECTION INPUT ; 
XYZ PIN BC
XYZ DIRECTION INPUT ;


MACRO GEN 
GEN PIN DECSEL_STG2[0]
GEN DIRECTION INPUT ; 
GEN PIN DECSEL_STG1[0]
GEN DIRECTION INPUT ;

---

这可以通过一个相对简单的

awk

脚本来完成，如下所示：

pax$ printf "MACRO XYZ\nline 2\nline3\n\nMACRO GEN\nline 6\n\n\n" | awk '
...$    $1=="MACRO"{mac=$2;print;next}$NF>0{print mac" "$0;next}{print}'
MACRO XYZ
XYZ line 2
XYZ line3
.
MACRO GEN
GEN line 6
.
. 

（那些<代码> ./COD>字符被我添加只是为了显示空白行在哪里）。 在脚本上展开：

$1 == "MACRO" {      # If first word is macro:
    mac=$2;          #    store second word,
    print;           #    print line,
    next             #    and go get next line.
}
$NF > 0 {            # Otherwise, if line has some fields:
    print mac" "$0;  #    print stored macro name before line,
    next             #    and go get next line.
}
{                    # Otherwise (lines with no fields):
    print            #    just print it as is.
}

---

akshay siddamal：：请在帖子中为您的示例输入/输出使用代码标签。