如何在使用awk的条件下删除csv文件行？

Question

我有3列的file.csv看起来像这样：

"/cw-0yLGNekxk4Vup/cwdd",-112.263,1
"001aCqAT2O+SbZfeC7UmhQ",,0
"/001DLIWmCvBwVCJrQjiQA",310,0
"fasdDLIWmCvBwVCJrQjiQA",0,0
...

但有些行没有第2列的值或0列2的值，我想删除这些行以获得：

"/cw-0yLGNekxk4Vup/cwdd",-112.263,1
"/001DLIWmCvBwVCJrQjiQA",310,0
...

如何在Ubuntu中使用awk做到这一点？

Answer 1

读错问题

在awk中删除行

awk -F, 'NF!=3' file

其实这样比较好

awk -F, '$2' file

假设$ 2可以是0

awk -F, '$2!=""' file
awk -F, '$2~/./' file

Answer 2

为什么不考虑使用grep ？

grep -v '^[^,],,'  file

Answer 3

您可以尝试以下awk命令，

$ awk '/",,/{next}{print}' file
"/cw-0yLGNekxk4Vup/cwdd",-112.263
"/001DLIWmCvBwVCJrQjiQA",310

要么

$ awk -F, 'NF>2{next;}{print}' file
"/cw-0yLGNekxk4Vup/cwdd",-112.263
"/001DLIWmCvBwVCJrQjiQA",310

要么

$ awk -F, 'NF<3' file
"/cw-0yLGNekxk4Vup/cwdd",-112.263
"/001DLIWmCvBwVCJrQjiQA",310