使用 awk 将列转换为行

Question

我有一个以下格式的输出文件。如何将其转换为第一列为 header 行，第二列为数据行的 csv？

输入：

"Node name","SLAVE_DB"
"Node ID","2"
"PostgreSQL version","10.17"
"Total data size","1 GB"
"Conninfo","host=192.168.0.1 port=5432 user=user dbname=postgres"
"Role","standby"
"WAL archiving","disabled (on standbys "archive_mode" must be set to "always" to be effective)"

期望的输出：

"Node name","Node ID","PostgreSQL version","Total data size","Conninfo","Role","WAL archiving"
"SLAVE_DB","2","10.17","1 GB","host=192.168.0.1 port=5432 user=user dbname=postgres","standby","disabled (on standbys ""archive_mode"" must be set to ""always"" to be effective)"

[编辑] 我通读了类似的问题并尝试了 awk -F "," '{ for (i=1;i<=NF;i++ ) printf $i ", " }'
但它不会产生所需的输出。我还想在数据中保留带引号的字段。

Answer 1

如果我们假设您的第一个字段（tag/name/header）中从来没有引号，那么在每个 Unix 机器上的任何 shell 中使用任何 awk：

$ cat tst.awk
BEGIN { FS=OFS="," }
{
    tags[NR] = 
    vals[NR] = 
}
END {
    for ( colNr=1; colNr<=NR; colNr++ ) {
        tag = tags[colNr]
        printf "%s%s", tag, (colNr<NR ? OFS : ORS)
    }
    for ( colNr=1; colNr<=NR; colNr++ ) {
        val = vals[colNr]
        gsub(/^"|"$/,"",val)
        gsub(/"/,"\"\"",val)
        printf "\"%s\"%s", val, (colNr<NR ? OFS : ORS)
    }
}

$ awk -f tst.awk file
"Node name","Node ID","PostgreSQL version","Total data size","Conninfo","Role","WAL archiving"
"SLAVE_DB","2","10.17","1 GB","host=192.168.0.1 port=5432 user=user dbname=postgres","standby","disabled (on standbys ""archive_mode"" must be set to ""always"" to be effective)"

如果您需要更多，请参阅。

使用 awk 将列转换为行

Convert columns to rows with awk

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