Grep：获取没有匹配行的所有文件

This may be a duplicate as I believe I already saw something like that but can't find it anymore so I'm asking :

这可能是重复的，因为我相信我已经看到了类似的东西但是找不到它所以我问：

I have a lots of files with multiple lines, and in most case, one of the line contain a certain patern. I would like to have every file that do not have a line with this patern.

我有很多具有多行的文件，在大多数情况下，其中一行包含某个patern。我想让每个文件都没有这个patern的行。

EDIT : Just need to use the "-L" grep option. This is a duplicate, I'm sure but don't find the original back ...

编辑：只需要使用“-L”grep选项。这是重复的，我敢肯定，但没有找到原来的背...

3 个解决方案

#1

The answer was just to get the "-L" option in order to have file WITHOUT the pattern ... Should have read the man more carefully!

答案只是为了获得“-L”选项以使文件没有模式......应该更仔细地阅读这个人！

#2

Grep returns 0/1 to indicate if there was a match, so you can do something like this:

Grep返回0/1表示是否匹配，所以你可以这样做：

for f in *.txt; do
    if ! grep -q "some expression" $f; then
        echo $f
    fi
done

EDIT: You can also use the -L option:

编辑：您还可以使用-L选项：

grep -L "some expression" *

grep -L“some expression”*

#3

try "count" and filter where equals ":0":

尝试“计数”并过滤等于“：0”：

grep -c [pattern] * | grep ":0"

(if you use TurboGREP cough, you won't have a -L switch ;))

（如果你使用TurboGREP咳嗽，你就不会有-L开关;））

#1