如何打開與給定正則表達式匹配最多的文件？

假設我有一個目錄~/mydir，裡面有一大堆文本文件。我想在這個目錄中搜尋searchterm，然後查看匹配最多的文件。我怎樣才能只使用一個命令來做到這一點？

將以下行放入腳本中即可：

grep -c "$1" ~/mydir/* | grep -v ':0' | sort -t: -k2 -r -n | head -1 | sed 's/:.*//' | xargs less

然後打電話./myscript searchterm

如果要遞歸搜尋，請在第一個命令中更改-c為。-cr``grep

此管道的各個部分，按順序：

grep -c "$1" ~/mydir/*    # Outputs a list of the files in ~/mydir/ with :<count>
                         # appended to each, where <count> is the number of
                         # matches of the $1 pattern in the file.

grep -v ':0'              # Removes the files that have 0 results from
                         # the list.

sort -t: -k2 -r -n        # Sorts the list in reverse numerical order
                         # (highest numbers first) based on the
                         # second ':'-delimited field

head -1                   # Extracts only the first result
                         # (the most matches)

sed 's/:.*//'             # Removes the trailing ':<count>' as we're
                         # done with it

xargs less                # Passes the resulting filename as
                         # an argument to less

如果根本沒有匹配，less 將打開空。

引用自：https://unix.stackexchange.com/questions/236519