bash打印仅前n个字符匹配的完整行

Question

I have created a sorted list of hashes for certain files

ffb01af8fda1e5c3b74d1eb384d021be1f1577c3 *./Pictures/camera/London 170713/P9110042.JPG
ffb01af8fda1e5c3b74d1eb384d021be1f1577c3 *./Pictures/london/P9110042.JPG

where there are duplicate hashes (just the hashes), I want to print the whole line of all matches

so say there where hashes ABC

A 1
B 2
B 3
C 4
C 5
C 6

in this example all the lines except the first one should be printed

B 2
B 3
C 4
C 5
C 6

Answer 1

Before you continue, look up fdupes .

If you don't want to use a robust tool specifically intended to find duplicate files, you can use sort | uniq sort | uniq :

$ cat file
A 1
B 2
B 3
C 4
C 5
C 6

$ sort file | uniq -w 1 -D
B 2
B 3
C 4
C 5
C 6

Answer 2

Using awk you can do (will work with unsorted file also):

awk 'FNR==NR{seen[$1]++; next} seen[$1]>1' file file
B 2
B 3
C 4
C 5
C 6