Question

我有一种压缩方法：

    (defn zip-project [project-id notebooks files]
  (with-open [out (ByteArrayOutputStream.)
              zip (ZipOutputStream. out)]
    (doseq [nb notebooks]
      (.putNextEntry zip (ZipEntry. (str "Notebooks/" (:notebook/name nb) ".bkr")))
      (let [nb-json (:notebook/contents nb)
            bytes (.getBytes nb-json)]
        (.write zip bytes))
      (.closeEntry zip))
    (doseq [{:keys [name content]} files]
      (.putNextEntry zip (ZipEntry. (str "Files/" name)))
      (io/copy content zip)
      (.closeEntry zip))
    (.finish zip)
    (.toByteArray out)))

在我制作一个zip之后，我想将它保存到像/tmp/sample/sample.zip这样的文件中，但我似乎无法做到。这就是我在做的事情：

(defn create-file! [path zip]
  (let [f (io/file path)]
    (io/make-parents f)
    (io/copy zip f)
    true))

问题是，当我从终端解压缩时，它说zip文件是空的，如果我使用Archive实用程序解压缩它，它会用cpgz扩展名提取。

我在这里做错了什么？

Answer 1

你基本上需要4件事

导入所有内容（通常使用(ns ...)，但您可以在repl中运行

(import 'java.io.FileOutputStream)
(import 'java.io.BufferedOutputStream)
(import 'java.io.ZipOutputStream)
(import 'java.util.zip.ZipOutputStream)
(import 'java.util.zip.ZipEntry)

您需要一种初始化流的方法。这可以通过->宏很好地完成：

(defn zip-stream
  "Opens a ZipOutputStream over the given file (as a string)"
  [file]
  (-> (FileOutputStream. file)
      BufferedOutputStream.
      ZipOutputStream.))

您需要一种方法来创建/关闭ZipOutputStream

中的条目

(defn create-zip-entry
  "Create a zip entry with the given name. That will be the name of the file inside the zipped file."
  [stream entry-name]
  (.putNextEntry stream (ZipEntry. entry-name)))

最后，您需要一种方法来撰写您的内容。

(defn write-to-zip
  "Writes a string to a zip stream as created by zip-stream"
  [stream str]
  (.write stream (.getBytes str)))

全部放在一起：

(with-open [stream (zip-stream "coolio.zip")]
  (create-zip-entry stream "foo1.txt")
  (write-to-zip stream "Hello Foo1")
  (.closeEntry stream) ;; don't forget to close entries
  (create-zip-entry stream "foo2.txt")
  (write-to-zip stream "Hello Foo 2")
  (.closeEntry stream))

结果：

将zip文件写入Clojure中的文件

1 个答案: