我有一个compojure应用程序,它使用环会话包装器来存储与当前用户关联的OAuth令牌.我希望这个令牌在服务器重新启动时保持可用,这样我就不必每次都要经过身份验证过程.
我假设使用cookie-store而不是默认的内存存储会有所帮助,但事实并非如此.我错过了什么？
这是代码的相关部分:
(defn auth-callback-handler
[session {code :code}]
(let [token (retrieve-token code)]
(-> (redirect "/") (assoc :session (assoc session :token token)))))
(defroutes app-routes
(GET "/" {session :session} (root-handler session))
(GET "/auth-callback" {session :session params :params} (auth-callback-handler session params))
(route/not-found "Not Found"))
(def app
(-> (handler/site app-routes)
(wrap-session {:store (cookie-store {:key "a 16-byte secret"})})))
该函数root-handler使用令牌来决定是否有人登录,但不会以会话信息的方式返回任何内容.
我想从String读取HTML,处理它并使用HXT将更改的文档作为String返回.由于此操作不需要IO,我宁愿执行箭头而runLA不是使用runX.
代码看起来像这样(省略处理以简化):
runLA (hread >>> writeDocumentToString [withOutputHTML, withIndent yes]) html
但是,html结果中缺少周围的标记:
["\n <head>\n <title>Bogus</title>\n </head>\n <body>\n Some trivial bogus text.\n </body>\n",""]
当我使用runX时,就像这样:
runX (readString [] html >>> writeDocumentToString [withOutputHTML, withIndent yes])
我得到了预期的结果:
["<html>\n <head>\n <title>Bogus</title>\n </head>\n <body>\n Some trivial bogus text.\n </body>\n</html>\n"]
为什么会这样,我该如何解决？
我实现了一个函数,它将给定输入集合的n-gram作为惰性seq返回.
(defn gen-ngrams
[n coll]
(if (>= (count coll) n)
(lazy-seq (cons (take n coll) (gen-ngrams n (rest coll))))))
当我用更大的输入集合调用此函数时,我希望看到执行时间的线性增加.但是,我观察到的时间比这更糟:
user> (time (count (gen-ngrams 3 (take 1000 corpus))))
"Elapsed time: 59.426 msecs"
998
user> (time (count (gen-ngrams 3 (take 10000 corpus))))
"Elapsed time: 5863.971 msecs"
9998
user> (time (count (gen-ngrams 3 (take 20000 corpus))))
"Elapsed time: 23584.226 msecs"
19998
user> (time (count (gen-ngrams 3 (take 30000 corpus))))
"Elapsed time: 54905.999 msecs"
29998
user> (time (count (gen-ngrams 3 (take 40000 corpus)))) …