bash - 如果空白，则将 CSV 字段复制到下一行（同一列） - Bash（最好）

Question

基本上（开玩笑地说）我有一个格式如下的 csv 文件：

"ID","Name","Phone Number"
"00001","Ricky Stallman","07771111111"
"00003","Harrison Ford","07701010101"
"00003","Harrison Ford",""
"00008","Bob Geldof","07712121212"

'Harrison Ford' 条目再次出现在我的 csv 中，旁边没有数字（这只是数据令人讨厌地呈现给我的方式）。我需要 csv 像这样读取（即，将上面行中的数字复制到下面的字段中）：

"ID","Name","Phone Number"
"00001","Ricky Stallman","07771111111"
"00003","Harrison Ford","07701010101"
"00003","Harrison Ford","07701010101"
"00008","Bob Geldof","07712121212"

如果有人有建议，最好在 Bash 中？

Answer 1

尝试这个：

awk -F',' '$3!~/""/{nbr=$3} {print $1","$2","nbr}' file

如果第三列是“”，则使用最后一个有效值。

Answer 2

可以使用 gawk 解决方案：

#!/usr/bin/gawk -f

match($0, /"([^\"]*)".*,"([^"]*)","([^"]*)"/, t) {
    key = t[1] "|" t[2]  ## Or just key = t[2] to be less strict.
    if (!(t[3] == "" && key in a)) {
        a[key] = t[3]
    }
    printf "\"%s\",\"%s\",\"\"%s\"\n", t[1], t[2], a[key]
}

浓缩：

gawk 'match($0, /"([^\"]*)".*,"([^"]*)","([^"]*)"/, t) { key = t[1] "|" t[2]; if (!(t[3] == "" && key in a)) a[key] = t[3]; printf "\"%s\",\"%s\",\"%s\"\n", t[1], t[2], a[key] }' file

输出：

"ID","Name","Phone Number"
"00001","Ricky Stallman","07771111111"
"00003","Harrison Ford","07701010101"
"00003","Harrison Ford","07701010101"
"00008","Bob Geldof","07712121212"