unix - 查找特定列是否具有特定字符串，如果该字符串存在则删除该列。

Question

如果文件的第 3 列包含字符串“BMR”，我想删除该文件，字符串可能是 - “BMR13234”，这意味着“BMR”是该字符串的静态部分，它可以包含任何其他字符。

因此，如果第 3 列包含 -“BMR*”（* 指任何字符），则应从文件中删除整个第 3 列。

例如： 输入：

"ABC","0.20","BMR1234442","2001-01-01"
"LMN","0.00","BMR2490289","2008-01-01"
"LTD","0.20","BMR2345577","2001-01-01"

输出应为：（仅当存在 BMR 时，否则不应删除第 3 列）

"ABC","0.20","2001-01-01"
"LMN","0.00","2008-01-01"
"LTD","0.20","2001-01-01"

请建议。

Answer 1

一种awk解决方案

awk -F, '$3~/^BMR/ {$3=x} {sub(/,,/,",")}1' OFS=, file
"ABC","0.20","2001-01-01"
"LMN","0.00","2008-01-01"
"LTD","0.20","2001-01-01"

测试是否3以 开头的字段BMR，如果是则将其设置为空白。
如果需要，Sub 删除额外的空字段。

Answer 2

sed 's/^\(\("[^"]*",\)\{2\}\)"BMR[^"]*",/\1/' YourFile

只要确保结构"blaba","blabla","BMR不包括之间的空间","