Question

我正在搜索几个日志，我想在最后一场比赛中上面和下面的几行进行比赛。

grep -A10 -B10 "searchString" my.log将在10行之后和之前打印所有匹配 grep "searchString" my.log | tail -n 1将打印最后一场比赛。

我希望将两者结合起来并获得最后一场比赛的10行之前和之后。

Answer 1

如果您想将所有命令都放在一个命令中，请尝试使用此awk

awk '/search/ {f=NR} {a[NR]=$0} END {while(i++<NR) if (i>f-3 && i<f+3) print a[i]}' file

工作原理：

awk '
/search/ {                      # Is pattern found
    f=NR}                       # yes, store the line number (it will then store only the last when all is run
    {
    a[NR]=$0}                   # Save all lines in an array "a"
END {
    while(i++<NR)               # Run trough all lines once more
        if (i>f-3 && i<f+3)     # If line number is +/- 2 compare to last found pattern, then 
            print a[i]          # Printe the line from the array "a"
    }' file                     # read the file

处理before和after

的更灵活的解决方案

awk '/fem/ {f=NR} {a[NR]=$0} END {while(i++<NR) if (i>=f-before && i<=f+after) print a[i]}' before=2 after=2 file

Answer 2

除非我完全误解了这个问题，否则你可以tail过去21行

 grep -A10 -B10  "searchString" my.log | tail -n 21

即

> for i in {1..10}; do echo $i >> example; done; \
echo foo >> example; \
for i in {1..10}; do echo $i >> example; done; \
for i in {A..Z}; do echo $i >> example; done; \
for i in {a..j}; do echo $i >> example; done; \
echo foo >> example; \
for i in {k..z}; do echo $i >> example; done
> grep -A10 -B10  foo example | tail -n 21
a
b
c
d
e
f
g
h
i
j
foo
k
l
m
n
o
p
q
r
s
t

Answer 3

你可以试试这个，

tac yourfile.log | grep -m1 -A2 -B2 'search' | tac

Grep在最后一场比赛之后和之前

3 个答案: