unix - 在特殊字段 AWK 中搜索字符

Question

我需要解析 awk 脚本中的字段来搜索特殊字符，如果存在则用“，”或“/”替换
awk 脚本将 CSV 转换为 DAT。定义的字段分隔符是；但有时用户会发送包含字段分隔符的评论。为了解决这个问题，我们需要解析 NUMBER $4 的 COMMENT 字段，如果该字段包含 ;用/或逗号代替
这是文件

"PAT";"TARO";"GEO";"COMMENT"
"FRT";"1256";"USA";"THIS IS A COMMENT ; AFTER COMMENT"

outcome expected

PAT TARO    GEO COMMENT
FRT 1256    USA THIS IS A COMMENT / AFTER COMMENT


BEGIN { 
  FS = ";" ;
  OFS = "   " ;

print "pat taro geo comment";   


}

NR==1{
next
}

{
pat= $1;
taro = $2;
geo = $3 ; 
comment = $4 ;
}

if $4 contains ";" then 
replace with "/"
end if;

{
    print "pat,taro,geo,comment";   

}

我怎样才能做到这一点 ？
先感谢您

Answer 1

如果字段中没有换行符，您可以使用例如 GNU awk 及其 FPAT特征:

$ gawk '
BEGIN {
    FPAT="([^;]*)|(\"[^\"]+\")"
}
{
    print $4
}' file

输出:

"COMMENT"
"THIS IS A COMMENT ; AFTER COMMENT"

如果您仍想更换 ;在评论中，添加 gsub(/;/,"/",$4)之前 print .
编辑 :

$ gawk '
BEGIN {
    FPAT="([^;]*)|(\"[^\"]+\")"   # FPAT;separates;semicolons;"and quotes"
    print "pat taro geo comment"  # print header
}
{
    for(i=1;i<=NF;i++)            # loop all 4 fields
        gsub(/^"|"$/,"",$i)       # remove quotes

    gsub(/;/,"/",$4)              # change the ; in $4 to /

    pat= $1                       # no need for this mut since you wanted
    taro = $2
    geo = $3
    comment = $4

    print pat,taro,geo,comment    # output new vars but you could as well:
    # print $1,$2,$3,$4           # use this too or
    # print $0                    # since record was rebuilt on gsub
}' file

输出:

pat taro geo comment
PAT TARO GEO COMMENT
FRT 1256 USA THIS IS A COMMENT / AFTER COMMENT