bash 删除 grep 输出中的重复项
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Removing duplicates in grep output
提问by john mas
I have a case where i got a results file with the following pattern:
我有一个案例，我得到了一个具有以下模式的结果文件：
path:pattern found
for example
例如
./user/home/file1:this is a game
in other words when i searched for some string i got the file and the line it found it.
换句话说，当我搜索某个字符串时，我得到了文件和找到它的行。
Problem is sometimes i have multiple cases in the same file so i would like to remove the duplicates files (the cases would be different so it's not possible).
问题是有时我在同一个文件中有多个案例，所以我想删除重复的文件（案例会有所不同，所以这是不可能的）。
Any help or ideas are appreciated :)
任何帮助或想法表示赞赏:)
End results is to turn this:
最终结果是变成这样：
/user/home/desktop/file1:this is a game
/user/home/desktop/file1:what kind of game
/user/home/desktop/file1:fast action game
into just the first results found without losing all the rest of the data in the file.
进入找到的第一个结果，而不会丢失文件中的所有其余数据。
Update1:
更新1：
So the actual file looks like this:
所以实际的文件是这样的：
/user/home/desktop/file1:this is a game
/user/home/desktop/file1:what kind of game
/user/home/desktop/file1:fast action game
/user/home/desktop/file2:a game
/user/home/desktop/file3:of game
/user/home/desktop/file4:fast game
i'm looking to get rid of the multiple occurrences in the same file so it should look like this:
我希望摆脱同一个文件中的多次出现，所以它应该是这样的：
/user/home/desktop/file1:this is a game
/user/home/desktop/file2:a game
/user/home/desktop/file3:of game
/user/home/desktop/file4:fast game
回答by codeforester
You could use sort -u:
你可以使用sort -u：
grep pattern files | sort -t: -u -k1,1
-t:- use : as the delimiter-k1,1- sort based on the first field only-u- removed duplicates (based on the first field)
-t:- 使用：作为分隔符-k1,1- 仅根据第一个字段排序-u- 删除重复项（基于第一个字段）
This will retain just one occurrence of files, removing any duplicates.
这将只保留一次文件，删除任何重复项。
For your example, this is the output you get:
对于您的示例，这是您获得的输出：
/user/home/desktop/file1:this is a game
In case you are looking for multiple distinct matches with a file, then:
如果您正在寻找与文件的多个不同匹配项，则：
grep pattern files | sort -u
