从没有破折号的字符串创建 UUID [英] Creating a UUID from a string with no dashes

问题描述

如何从没有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

Clojure 的 #uuid 标记文字 是java.util.UUID/fromString.并且，fromString 将它用-"分割并将其转换为两个 Long 值.(UUID 的格式被标准化为 8-4-4-4-12 十六进制数字，但-"实际上仅用于验证和视觉识别.)

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")))

如果你想要没有正则表达式的东西，你可以使用 ByteBuffer 和 DatatypeConverter.

If you want something without regular expressions, you can use a ByteBuffer and 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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