linux - Grep 包含多个元字符的字符串并在匹配后提取 3 行

Question

我想在文件中搜索 1:N:0:CGATGT 并提取包含 1:N:0:CGATGT 的行和另外 3 行之后(每场比赛总共 4 行)。我已经厌倦了 grep 多种方式，但都失败了:

[ssabri@login2 data]$ history | tail -n 8
 1028  zcat A1_S1_L008_R1_001.fastq.gz | grep -A4 "1[[:]][[N]][[:]]0[[:]]CGATGT" | wc -l
 1029  zcat A1_S1_L008_R1_001.fastq.gz | grep -A4 "1[[:]][[N]][[:]]0[[:]]CGATGT$" | wc -l
 1030  zcat A1_S1_L008_R1_001.fastq.gz | grep -A4 "1[[:]][[N]][[:]][[0]][[:]]CGATGT$" | wc -l
 1031  zcat A1_S1_L008_R1_001.fastq.gz | grep -A4 -w "1[[:]][[N]][[:]][[0]][[:]]CGATGT$" | wc -l
 1032  zcat A1_S1_L008_R1_001.fastq.gz | egrep -A4 -w "1[[:]][[N]][[:]][[0]][[:]]CGATGT$" | wc -l
 1033  zcat A1_S1_L008_R1_001.fastq.gz | grep -x -A4 -w "1:N:0:CGATGT" | wc -l
 1034  zcat A1_S1_L008_R1_001.fastq.gz | grep -E -A4 -w "1:N:0:CGATGT" | wc -l
 1035  zcat A1_S1_L008_R1_001.fastq.gz | grep -A4 -w "1\:N\:0\:CGATGT$" | wc -l

编辑:输入文件看起来像这样:

[ssabri@login2 data]$ zcat A1_S1_L008_R1_001.fastq.gz | head -n 12
@J00153:28:H7LNWBBXX:8:1101:28625:1191 1:N:0:CGAGGT
ACNTGCTCCATCCATAGCACCTAGAACAGAGCCTGGNACAGAANAAGNGC
+
A-#<-<<FJJAJFFFF-FJJJJJAJFJJJFF-A-FA#JJJJFJ#JJA#FJ
@J00153:28:H7LNWBBXX:8:1101:29457:1191 1:N:0:CGATGT
GTNGTGGTAGATCTGGACGCGGCTGAAGGCCTGGGGNCCCGTGNCAGN
+
-<#<FJJJJJJJJJJJJJJJJJJJJJJFJJJJJJJJ#JJJJJJ#JJJ#
@J00153:28:H7LNWBBXX:8:1101:31000:1191 1:N:0:CCATGT
TCNAATTATCACCATTACAGGAGGGTCAGTAGAACANGCGTTCTGGTNGG
+
<A#<AFFJJJFJJJFJJJJJJFFFJ7A<<JJFJJJJ#JJJAFJJJJF#-A

Answer 1

grep -A3 "1:N:0:CGATGT" file
@J00153:28:H7LNWBBXX:8:1101:29457:1191 1:N:0:CGATGT
GTNGTGGTAGATCTGGACGCGGCTGAAGGCCTGGGGNCCCGTGNCAGN
+
-<#<FJJJJJJJJJJJJJJJJJJJJJJFJJJJJJJJ#JJJJJJ#JJJ#

有时更简单的想法更好，在这里你不需要任何正则表达式扩展，因为你是字符串匹配而没有任何需要转义的特殊正则表达式字符。 A(after) 上下文匹配应为 3，因为您需要 3 条尾随行(总计为 4 条匹配行)。