从不带破折号的字符串创建UUID [英] Creating a UUID from a string with no dashes

问题描述

如何从不带破折号的字符串创建java.util.UUID？

How would I create a java.util.UUID from a string with no dashes?

"5231b533ba17478798a3f2df37de2aD7" => #uuid "5231b533-ba17-4787-98a3-f2df37de2aD7"

推荐答案

p> Clojure的 #uuid 标记的文字是传递到 java.util.UUID / fromString 。并且， fromString 将它分割为 - ，并将其转换为两个 Long 值。 （ UUID 的格式标准化为8-4-4-4-12十六进制数字，但 - 为）

Clojure's #uuid tagged literal is a pass-through to java.util.UUID/fromString. And, fromString splits it by the "-" and converts it into two Long values. (The format for UUID is standardized to 8-4-4-4-12 hex digits, but the "-" are really only there for validation and visual identification.)

直接的解决方案是重新插入 - 并使用 java.util.UUID / fromString

The straight forward solution is to reinsert the "-" and use java.util.UUID/fromString.

(defn uuid-from-string [data]
  (java.util.UUID/fromString
   (clojure.string/replace data
                           #"(\w{8})(\w{4})(\w{4})(\w{4})(\w{12})"
                           "$1-$2-$3-$4-$5")))

b $ b

如果您想要没有正则表达式的内容，可以使用 ByteBuffer 和 DatatypeConverter 。

(defn uuid-from-string [data]
  (let [buffer (java.nio.ByteBuffer/wrap 
                 (javax.xml.bind.DatatypeConverter/parseHexBinary data))]
    (java.util.UUID. (.getLong buffer) (.getLong buffer))))

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