我有一个包含大量（3400多行）网址列表的文件，每个网址都是这样的：
http://examplesite/rn/index.php?PageID=SF01_02_01&ID=2015-12-23-0012
我想要做的是使用sed选择包含今天日期的所有行。
我一直在尝试的是：
datetoday=$(date +%Y-%m-%d)
sed "/$datetoday/!d" en.html
然而，这似乎使文件空白，表明没有行符合字符串。有什么建议吗？
答案 0 :(得分：2)
您可以使用标记type Traversal s t a b = forall f. Applicative f => (a -> f b) -> (s -> f t)
data TraversalR s t a b = TraversalR {
toListOfR :: s -> [a],
setListR :: [b] -> s -> t
}
newtype CL a b = CL { getCL :: [a] } -- ConstantList
instance Functor (CL a) where
fmap _ (CL xs) = CL xs
instance Applicative (CL a) where
pure _ = CL []
(CL xs) <*> (CL ys) = CL (xs ++ ys)
collectBs :: State [b] b
collectBs = state $ \xs -> case xs of [] -> error "Too few bs"
(y:ys) -> (y,ys)
travToTravR :: Traversal s t a b -> TraversalR s t a b
travToTravR tr = TraversalR {
toListOfR = getCL . tr (CL . pure),
setListR = \bs s -> evalState (tr (const collectBs) s) bs
}
travRToTrav :: TraversalR s t a b -> Traversal s t a b
travRToTrav trR a2fb s = (`setL` s) <$> f_bs
where as = toListOfR trR s
f_bs = sequenceA . map a2fb $ as
setL = setListR trR
代替p：
d
答案 1 :(得分：1)
您可以使用GNU awk而不是2个单独的命令：
awk '$0~strftime("%F")'