当第 28 列中没有数据时使用 bash (awk/sed) 删除行

Question

下面是我的 .csv 文件的简要示例：

"X","Y","Z","N"
"X","Y","Z","N"
"X","Y","","N"
"X","Y","Z","N"
"X","Y","Z","N"
"X","Y","","N"
"X","Y","Z","N"
"X","Y","Z","N"
"X","Y","","N"

在第三列（“Z”列）中，我缺少单元格（第 3,6 和 9 行）。最好使用 awk 或 sed，我想专门针对第 3 列，如果任何单元格为空白，我想删除整行。我的最终结果是：

"X","Y","Z","N"
"X","Y","Z","N"
"X","Y","Z","N"
"X","Y","Z","N"
"X","Y","Z","N"
"X","Y","Z","N"

对于我的实际项目，下面是我的一个文件的精确副本 - https://github.com/drphillgood/riotapidata/blob/master/csv/game3.csv。您将在第 28 列 (participants__participantId) 中看到只有某些单元格有数据（最后一列 Participants__playerName 相同）。如果此列中的一个单元格为空，我想使用 .sh 脚本删除整行。最终文件会像这样 - https://github.com/drphillgood/riotapidata/blob/master/csv/game3_v2.csv

Answer 1

更简单的 AWK 命令：

awk -F , '$3 != "\"\"" {print}' inputfile > outputfile

将字段分隔符设置为逗号并打印第三个字段仅包含 的每一行""。

这不足以处理字段中包含逗号的 CSV 文件。它期望空字段由空引号组成。