Question

基本上（稍微说一下）我有一个csv文件，格式如下：

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

“哈里森福特”条目再次出现在我的csv中，旁边没有数字（这只是数据令我烦恼的方式）。我需要csv这样读（即将上面一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"

最好在Bash中是否有人有建议？

Answer 1

试试这个：

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]
}

冷凝的：

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"

如果空白，则将CSV字段复制到下一行（同一列） - Bash（最好）

2 个答案: