Linux - 如何根据字段值从文件中删除某些行

Question

I want to remove certain lines from a tab-delimited file and write output to a new file.

a   b   c   2017-09-20
a   b   c   2017-09-19
es  fda d   2017-09-20
es  fda d   2017-09-19

The 4th column is Date, basically I want to keep only lines that has 4th column as "2017-09-19" (keep line 2&4) and write to a new file. The new file should have same format as the raw file.

How to write the linux command for this example?

Note : The search criteria should be on the 4th field as I have other fields in the real data and possibly have same value as 4th field.

Answer 1

Use grep to filter:

cat file.txt | grep '2017-09-19' > filtered_file.txt

This is not perfect, since the string 2017-09-19 is not required to appear in the 4th column, but if your file looks like the example, it'll work.

Answer 2

With awk:

awk 'BEGIN{OFS="\t"} $4=="2017-09-19"' file

OFS : output field separator, a space by default

Answer 3

Sed solution:

sed -nr "/^([^\t]*\t){3}2017-09-19/p" input.txt >output.txt

this is:

• -n - don't output every line
• -r - extended regular expresion
• /regexp/p - print line that contains regular expression regexp
• ^ - begin of line
• (regexp){3} - repeat regexp 3 times
• [^\\t] - any character except tab
• \\t - tab character
• * - repeat characters multiple times
• 2017-09-19 - search text

That is, skip 3 columns separated by a tab from the beginning of the line, and then check that the value of column 4 coincides with the required value.

Answer 4

awk '/2017-09-19/' file >newfile

cat newfile
a   b   c   2017-09-19
es  fda d   2017-09-19