在 JavaScript 中，有許多好的（也有壞的）方法可以將字串拆分為陣列。
例如，在 ES6 中只使用擴展運算子：
let str = "example 1";
let arr = [...str];
console.log(arr);
// ['e', 'x', 'a', 'm', 'p', 'l', 'e', ' ', '1']
當包含表情符號時，其中很多仍然有效
let str = "example 2 ??";
let arr = [...str];
console.log(arr);
// ['e', 'x', 'a', 'm', 'p', 'l', 'e', ' ', '2', ' ', '??']
但是，如果字串中包含復合表情符號，則到目前為止我找到的所有解決方案（例如1、2）都會失敗
let str = "example 3 ???♀? ???????????";
let arr = [...str];
console.log(arr);
// ['e', 'x', 'a', 'm', 'p', 'l', 'e', ' ', '3', ' ', '??', '?', '♀', '?', ' ', '??', '?', '??', '?', '??', '?', '??']
有grapheme-splitter，但這已經多年沒有更新了。另外，我更喜歡找到一種不需要包含外部庫/包的方法。
這種方法檢測（許多，不是全部）復合表情符號是正確的，但不會分離文本部分。（我還不明白它是如何作業的，所以我可以調整它。）
let str = "example 3 ???♀? ???????????";
let regex = /([\uD800-\uDBFF][\uDC00-\uDFFF](?:[\u200D\uFE0F][\uD800-\uDBFF][\uDC00-\uDFFF]){2,}|\uD83D\uDC69(?:\u200D(?:(?:\uD83D\uDC69\u200D)?\uD83D\uDC67|(?:\uD83D\uDC69\u200D)?\uD83D\uDC66)|\uD83C[\uDFFB-\uDFFF])|\uD83D\uDC69\u200D(?:\uD83D\uDC69\u200D)?\uD83D\uDC66\u200D\uD83D\uDC66|\uD83D\uDC69\u200D(?:\uD83D\uDC69\u200D)?\uD83D\uDC67\u200D(?:\uD83D[\uDC66\uDC67])|\uD83C\uDFF3\uFE0F\u200D\uD83C\uDF08|(?:\uD83C[\uDFC3\uDFC4\uDFCA]|\uD83D[\uDC6E\uDC71\uDC73\uDC77\uDC81\uDC82\uDC86\uDC87\uDE45-\uDE47\uDE4B\uDE4D\uDE4E\uDEA3\uDEB4-\uDEB6]|\uD83E[\uDD26\uDD37-\uDD39\uDD3D\uDD3E\uDDD6-\uDDDD])(?:\uD83C[\uDFFB-\uDFFF])\u200D[\u2640\u2642]\uFE0F|\uD83D\uDC69(?:\uD83C[\uDFFB-\uDFFF])\u200D(?:\uD83C[\uDF3E\uDF73\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92])|(?:\uD83C[\uDFC3\uDFC4\uDFCA]|\uD83D[\uDC6E\uDC6F\uDC71\uDC73\uDC77\uDC81\uDC82\uDC86\uDC87\uDE45-\uDE47\uDE4B\uDE4D\uDE4E\uDEA3\uDEB4-\uDEB6]|\uD83E[\uDD26\uDD37-\uDD39\uDD3C-\uDD3E\uDDD6-\uDDDF])\u200D[\u2640\u2642]\uFE0F|\uD83C\uDDFD\uD83C\uDDF0|\uD83C\uDDF6\uD83C\uDDE6|\uD83C\uDDF4\uD83C\uDDF2|\uD83C\uDDE9(?:\uD83C[\uDDEA\uDDEC\uDDEF\uDDF0\uDDF2\uDDF4\uDDFF])|\uD83C\uDDF7(?:\uD83C[\uDDEA\uDDF4\uDDF8\uDDFA\uDDFC])|\uD83C\uDDE8(?:\uD83C[\uDDE6\uDDE8\uDDE9\uDDEB-\uDDEE\uDDF0-\uDDF5\uDDF7\uDDFA-\uDDFF])|(?:\u26F9|\uD83C[\uDFCB\uDFCC]|\uD83D\uDD75)(?:\uFE0F\u200D[\u2640\u2642]|(?:\uD83C[\uDFFB-\uDFFF])\u200D[\u2640\u2642])\uFE0F|(?:\uD83D\uDC41\uFE0F\u200D\uD83D\uDDE8|\uD83D\uDC69(?:\uD83C[\uDFFB-\uDFFF])\u200D[\u2695\u2696\u2708]|\uD83D\uDC69\u200D[\u2695\u2696\u2708]|\uD83D\uDC68(?:(?:\uD83C[\uDFFB-\uDFFF])\u200D[\u2695\u2696\u2708]|\u200D[\u2695\u2696\u2708]))\uFE0F|\uD83C\uDDF2(?:\uD83C[\uDDE6\uDDE8-\uDDED\uDDF0-\uDDFF])|\uD83D\uDC69\u200D(?:\uD83C[\uDF3E\uDF73\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]|\u2764\uFE0F\u200D(?:\uD83D\uDC8B\u200D(?:\uD83D[\uDC68\uDC69])|\uD83D[\uDC68\uDC69]))|\uD83C\uDDF1(?:\uD83C[\uDDE6-\uDDE8\uDDEE\uDDF0\uDDF7-\uDDFB\uDDFE])|\uD83C\uDDEF(?:\uD83C[\uDDEA\uDDF2\uDDF4\uDDF5])|\uD83C\uDDED(?:\uD83C[\uDDF0\uDDF2\uDDF3\uDDF7\uDDF9\uDDFA])|\uD83C\uDDEB(?:\uD83C[\uDDEE-\uDDF0\uDDF2\uDDF4\uDDF7])|[#\*0-9]\uFE0F\u20E3|\uD83C\uDDE7(?:\uD83C[\uDDE6\uDDE7\uDDE9-\uDDEF\uDDF1-\uDDF4\uDDF6-\uDDF9\uDDFB\uDDFC\uDDFE\uDDFF])|\uD83C\uDDE6(?:\uD83C[\uDDE8-\uDDEC\uDDEE\uDDF1\uDDF2\uDDF4\uDDF6-\uDDFA\uDDFC\uDDFD\uDDFF])|\uD83C\uDDFF(?:\uD83C[\uDDE6\uDDF2\uDDFC])|\uD83C\uDDF5(?:\uD83C[\uDDE6\uDDEA-\uDDED\uDDF0-\uDDF3\uDDF7-\uDDF9\uDDFC\uDDFE])|\uD83C\uDDFB(?:\uD83C[\uDDE6\uDDE8\uDDEA\uDDEC\uDDEE\uDDF3\uDDFA])|\uD83C\uDDF3(?:\uD83C[\uDDE6\uDDE8\uDDEA-\uDDEC\uDDEE\uDDF1\uDDF4\uDDF5\uDDF7\uDDFA\uDDFF])|\uD83C\uDFF4\uDB40\uDC67\uDB40\uDC62(?:\uDB40\uDC77\uDB40\uDC6C\uDB40\uDC73|\uDB40\uDC73\uDB40\uDC63\uDB40\uDC74|\uDB40\uDC65\uDB40\uDC6E\uDB40\uDC67)\uDB40\uDC7F|\uD83D\uDC68(?:\u200D(?:\u2764\uFE0F\u200D(?:\uD83D\uDC8B\u200D)?\uD83D\uDC68|(?:(?:\uD83D[\uDC68\uDC69])\u200D)?\uD83D\uDC66\u200D\uD83D\uDC66|(?:(?:\uD83D[\uDC68\uDC69])\u200D)?\uD83D\uDC67\u200D(?:\uD83D[\uDC66\uDC67])|\uD83C[\uDF3E\uDF73\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92])|(?:\uD83C[\uDFFB-\uDFFF])\u200D(?:\uD83C[\uDF3E\uDF73\uDF93\uDFA4\uDFA8\uDFEB\uDFED]|\uD83D[\uDCBB\uDCBC\uDD27\uDD2C\uDE80\uDE92]))|\uD83C\uDDF8(?:\uD83C[\uDDE6-\uDDEA\uDDEC-\uDDF4\uDDF7-\uDDF9\uDDFB\uDDFD-\uDDFF])|\uD83C\uDDF0(?:\uD83C[\uDDEA\uDDEC-\uDDEE\uDDF2\uDDF3\uDDF5\uDDF7\uDDFC\uDDFE\uDDFF])|\uD83C\uDDFE(?:\uD83C[\uDDEA\uDDF9])|\uD83C\uDDEE(?:\uD83C[\uDDE8-\uDDEA\uDDF1-\uDDF4\uDDF6-\uDDF9])|\uD83C\uDDF9(?:\uD83C[\uDDE6\uDDE8\uDDE9\uDDEB-\uDDED\uDDEF-\uDDF4\uDDF7\uDDF9\uDDFB\uDDFC\uDDFF])|\uD83C\uDDEC(?:\uD83C[\uDDE6\uDDE7\uDDE9-\uDDEE\uDDF1-\uDDF3\uDDF5-\uDDFA\uDDFC\uDDFE])|\uD83C\uDDFA(?:\uD83C[\uDDE6\uDDEC\uDDF2\uDDF3\uDDF8\uDDFE\uDDFF])|\uD83C\uDDEA(?:\uD83C[\uDDE6\uDDE8\uDDEA\uDDEC\uDDED\uDDF7-\uDDFA])|\uD83C\uDDFC(?:\uD83C[\uDDEB\uDDF8])|(?:\u26F9|\uD83C[\uDFCB\uDFCC]|\uD83D\uDD75)(?:\uD83C[\uDFFB-\uDFFF])|(?:\uD83C[\uDFC3\uDFC4\uDFCA]|\uD83D[\uDC6E\uDC71\uDC73\uDC77\uDC81\uDC82\uDC86\uDC87\uDE45-\uDE47\uDE4B\uDE4D\uDE4E\uDEA3\uDEB4-\uDEB6]|\uD83E[\uDD26\uDD37-\uDD39\uDD3D\uDD3E\uDDD6-\uDDDD])(?:\uD83C[\uDFFB-\uDFFF])|(?:[\u261D\u270A-\u270D]|\uD83C[\uDF85\uDFC2\uDFC7]|\uD83D[\uDC42\uDC43\uDC46-\uDC50\uDC66\uDC67\uDC70\uDC72\uDC74-\uDC76\uDC78\uDC7C\uDC83\uDC85\uDCAA\uDD74\uDD7A\uDD90\uDD95\uDD96\uDE4C\uDE4F\uDEC0\uDECC]|\uD83E[\uDD18-\uDD1C\uDD1E\uDD1F\uDD30-\uDD36\uDDD1-\uDDD5])(?:\uD83C[\uDFFB-\uDFFF])|\uD83D\uDC68(?:\u200D(?:(?:(?:\uD83D[\uDC68\uDC69])\u200D)?\uD83D\uDC67|(?:(?:\uD83D[\uDC68\uDC69])\u200D)?\uD83D\uDC66)|\uD83C[\uDFFB-\uDFFF])|(?:[\u261D\u26F9\u270A-\u270D]|\uD83C[\uDF85\uDFC2-\uDFC4\uDFC7\uDFCA-\uDFCC]|\uD83D[\uDC42\uDC43\uDC46-\uDC50\uDC66-\uDC69\uDC6E\uDC70-\uDC78\uDC7C\uDC81-\uDC83\uDC85-\uDC87\uDCAA\uDD74\uDD75\uDD7A\uDD90\uDD95\uDD96\uDE45-\uDE47\uDE4B-\uDE4F\uDEA3\uDEB4-\uDEB6\uDEC0\uDECC]|\uD83E[\uDD18-\uDD1C\uDD1E\uDD1F\uDD26\uDD30-\uDD39\uDD3D\uDD3E\uDDD1-\uDDDD])(?:\uD83C[\uDFFB-\uDFFF])?|(?:[\u231A\u231B\u23E9-\u23EC\u23F0\u23F3\u25FD\u25FE\u2614\u2615\u2648-\u2653\u267F\u2693\u26A1\u26AA\u26AB\u26BD\u26BE\u26C4\u26C5\u26CE\u26D4\u26EA\u26F2\u26F3\u26F5\u26FA\u26FD\u2705\u270A\u270B\u2728\u274C\u274E\u2753-\u2755\u2757\u2795-\u2797\u27B0\u27BF\u2B1B\u2B1C\u2B50\u2B55]|\uD83C[\uDC04\uDCCF\uDD8E\uDD91-\uDD9A\uDDE6-\uDDFF\uDE01\uDE1A\uDE2F\uDE32-\uDE36\uDE38-\uDE3A\uDE50\uDE51\uDF00-\uDF20\uDF2D-\uDF35\uDF37-\uDF7C\uDF7E-\uDF93\uDFA0-\uDFCA\uDFCF-\uDFD3\uDFE0-\uDFF0\uDFF4\uDFF8-\uDFFF]|\uD83D[\uDC00-\uDC3E\uDC40\uDC42-\uDCFC\uDCFF-\uDD3D\uDD4B-\uDD4E\uDD50-\uDD67\uDD7A\uDD95\uDD96\uDDA4\uDDFB-\uDE4F\uDE80-\uDEC5\uDECC\uDED0-\uDED2\uDEEB\uDEEC\uDEF4-\uDEF8]|\uD83E[\uDD10-\uDD3A\uDD3C-\uDD3E\uDD40-\uDD45\uDD47-\uDD4C\uDD50-\uDD6B\uDD80-\uDD97\uDDC0\uDDD0-\uDDE6])|(?:[#\*0-9\xA9\xAE\u203C\u2049\u2122\u2139\u2194-\u2199\u21A9\u21AA\u231A\u231B\u2328\u23CF\u23E9-\u23F3\u23F8-\u23FA\u24C2\u25AA\u25AB\u25B6\u25C0\u25FB-\u25FE\u2600-\u2604\u260E\u2611\u2614\u2615\u2618\u261D\u2620\u2622\u2623\u2626\u262A\u262E\u262F\u2638-\u263A\u2640\u2642\u2648-\u2653\u2660\u2663\u2665\u2666\u2668\u267B\u267F\u2692-\u2697\u2699\u269B\u269C\u26A0\u26A1\u26AA\u26AB\u26B0\u26B1\u26BD\u26BE\u26C4\u26C5\u26C8\u26CE\u26CF\u26D1\u26D3\u26D4\u26E9\u26EA\u26F0-\u26F5\u26F7-\u26FA\u26FD\u2702\u2705\u2708-\u270D\u270F\u2712\u2714\u2716\u271D\u2721\u2728\u2733\u2734\u2744\u2747\u274C\u274E\u2753-\u2755\u2757\u2763\u2764\u2795-\u2797\u27A1\u27B0\u27BF\u2934\u2935\u2B05-\u2B07\u2B1B\u2B1C\u2B50\u2B55\u3030\u303D\u3297\u3299]|\uD83C[\uDC04\uDCCF\uDD70\uDD71\uDD7E\uDD7F\uDD8E\uDD91-\uDD9A\uDDE6-\uDDFF\uDE01\uDE02\uDE1A\uDE2F\uDE32-\uDE3A\uDE50\uDE51\uDF00-\uDF21\uDF24-\uDF93\uDF96\uDF97\uDF99-\uDF9B\uDF9E-\uDFF0\uDFF3-\uDFF5\uDFF7-\uDFFF]|\uD83D[\uDC00-\uDCFD\uDCFF-\uDD3D\uDD49-\uDD4E\uDD50-\uDD67\uDD6F\uDD70\uDD73-\uDD7A\uDD87\uDD8A-\uDD8D\uDD90\uDD95\uDD96\uDDA4\uDDA5\uDDA8\uDDB1\uDDB2\uDDBC\uDDC2-\uDDC4\uDDD1-\uDDD3\uDDDC-\uDDDE\uDDE1\uDDE3\uDDE8\uDDEF\uDDF3\uDDFA-\uDE4F\uDE80-\uDEC5\uDECB-\uDED2\uDEE0-\uDEE5\uDEE9\uDEEB\uDEEC\uDEF0\uDEF3-\uDEF8]|\uD83E[\uDD10-\uDD3A\uDD3C-\uDD3E\uDD40-\uDD45\uDD47-\uDD4C\uDD50-\uDD6B\uDD80-\uDD97\uDDC0\uDDD0-\uDDE6])\uFE0F)/;
let arr = str.split(regex).filter(Boolean);
console.log(arr); // ['example 3 ', '???♀?', ' ', '???????????']
恕我直言，它非常“丑陋”。也許這可以通過使用表情符號屬性變得更漂亮？
由于表情符號現在無處不在，我覺得必須有一種簡單的方法可以將帶有復合表情符號的字串拆分為陣列。
uj5u.com熱心網友回復：
對此沒有簡短的解決方案代碼，因為 Unicode 變得越來越復雜。您可以查看unicode.org本身提供的EBNF 和 Regex。在撰寫本文時，它寫道：
正則運算式
\p{RI} \p{RI} | \p{Emoji} ( \p{EMod} | \x{FE0F} \x{20E3}? | [\x{E0020}-\x{E007E}] \x{E007F} )? (\x{200D} \p{Emoji} ( \p{EMod} | \x{FE0F} \x{20E3}? | [\x{E0020}-\x{E007E}] \x{E007F} )? )*
轉換為 JavaScript，正則運算式為：
/\p{RI}\p{RI}|\p{Emoji}(\p{EMod}|\u{FE0F}\u{20E3}?|[\u{E0020}-\u{E007E}] \u{E007F})?(\u{200D}\p{Emoji}(\p{EMod}|\u{FE0F}\u{20E3}?|[\u{E0020}-\u{E007E}] \u{E007F})?)*/gu
如果您還想匹配任何其他字符，則只需附加|.到正則運算式，并添加s修飾符，以便點通配符也匹配換行符。
使用您的輸入示例：
let regex = /\p{RI}\p{RI}|\p{Emoji}(\p{EMod}|\u{FE0F}\u{20E3}?|[\u{E0020}-\u{E007E}] \u{E007F})?(\u{200D}\p{Emoji}(\p{EMod}|\u{FE0F}\u{20E3}?|[\u{E0020}-\u{E007E}] \u{E007F})?)*|./gus;
let str = "example 3 ???♀? ???????????";
document.write('<pre>' str.match(regex).join('\n'));注意：我不得不使用document.write而不是在console.log這里，因為consoleStack Snippets 提供的實作不能勝任這項任務。
現在，對這個正則運算式有一些批評，這導致了許多不同的風格。例如，在這個答案中，提出了以下改編和擴展來處理一些問題：
/\p{RI}\p{RI}|\p{Emoji}(\p{EMod} |\u{FE0F}\u{20E3}?|[\u{E0020}-\u{E007E}] \u{E007F})?(\u{200D}\p{Emoji}(\p{EMod} |\u{FE0F}\u{20E3}?|[\u{E0020}-\u{E007E}] \u{E007F})?) |\p{EPres}(\p{EMod} |\u{FE0F}\u{20E3}?|[\u{E0020}-\u{E007E}] \u{E007F})?|\p{Emoji}(\p{EMod} |\u{FE0F}\u{20E3}?|[\u{E0020}-\u{E007E}] \u{E007F})/gu
同樣，您可以附加|.到正則運算式（并添加s修飾符）以使其也匹配普通字符。
轉載請註明出處，本文鏈接：https://www.uj5u.com/gongcheng/399290.html
標籤：javascript 数组 细绳 表情符号