bash中Mth列中字符串N个字符的条件

Question

我有一个样本

$ cat c.csv
a,1234543,c
b,1231456,d
c,1230654,e

我只需要在第二列的第四个字符而不是0或1的情况下进行grep。

输出必须是

a,1234543,c

我只知道

awk -F, 'BEGIN { OFS = FS } $2 ~/^[2-9]/' c.csv

有可能在第四个角色上加上条件吗？

Answer 1

你能试一下吗。

awk 'BEGIN{FS=","} substr($2,4,1)!=0 && substr($2,4,1)!=1' Input_file

或根据教育署网站的建议：

awk 'BEGIN{FS=","} substr($2,4,1)!~[01]' Input_file

解释：在这里添加了对上述代码的详细说明。

awk '                                        ##Starting awk program from here.
BEGIN{                                       ##Starting BEGIN section from here.
  FS=","                                     ##Setting field separator as comma here.
}                                            ##Closing BLOCK for this program BEGIN section.
substr($2,4,1)!=0 && substr($2,4,1)!=1       ##Checking conditions if 4th character of current line is NOT 0 and 1 then print the current line.
' Input_file                                 ##Mentioning Input_file name here.

Answer 2

这可能适用于您(GNU或grep)：

grep -vE '^([^,]*,){1}[^,]{3}[01]' file

或者：

sed -E '/^([^,]*,){1}[^,]{3}[01]/d' file  

替换该列中m‘-1列的1和n’-1字符的3。

Answer 3

Grep是答案。但是下面是另一种使用数组和变量替换的方法

test=( $(cat c.csv) ) # load c.csv data to an array
echo ${test[@]//*,???[0-1]*/} # print all items from an array,
# but remove the ones that correspond to this regex *,???[0-1]*
# so 'b,1231456,d' and 'c,1230654,e' from example will be removed
# and only 'a,1234543,c' will be printed