Javascript 正则表达式：测试人名

Javascript regex: test people's name

提问by yoda

From this questions : javascript regex : only english letters allowed

从这个问题：javascript正则表达式：只允许英文字母

How can I make that expression test for people's name? Currently it doesn't allow spaces between names at all. I need to be able to match something like John Doe

我怎样才能对人们的名字进行表达测试？目前它根本不允许名称之间有空格。我需要能够匹配像John Doe这样的东西

Cheers

干杯

回答by meder omuraliev

let result = /^[a-zA-Z ]+$/.test( 'John Doe');
console.log(result);

Throw any symbols you need in the character class. This is whyI said be specificabout exactlywhat you want to validate. This regex will not account for accented characters, if you care about that you'd most likely better go with unicode matching.

在字符类中抛出您需要的任何符号。这就是为什么我说的是具体的关于究竟要验证什么。此正则表达式不会考虑重音字符，如果您关心这一点，您很可能最好使用 unicode 匹配。

回答by user2340939

^\s*([A-Za-z]{1,}([\.,] |[-']| ))+[A-Za-z]+\.?\s*$

Similar as @Stephen_Wylie's solution, but shorter (better?).

类似于@Stephen_Wylie 的解决方案，但更短（更好？）。

回答by andyengle

FWIW, I wrote this regex for cleaning up names:

FWIW，我写了这个正则表达式来清理名称：

let name = "John Doe";
let result = name.replace(/[^A-Za-z0-9_'-]/gi, '');
console.log(result);

This will let names like O'Brien, Smith, Smith-O'Brian pass through, but will prevent "names" like Engle;drop user; from being passed through. Anyway, it's four years after the fact, but hopefully my answer is of some use.

这将让像 O'Brien、Smith、Smith-O'Brian 这样的名字通过，但会阻止像 Engle 这样的“名字”；drop user; 从被通过。无论如何，事实已经过去了四年，但希望我的回答有些用处。

回答by Stephen Wylie

Try this:

尝试这个：

/^(([A-Za-z]+[\-\']?)*([A-Za-z]+)?\s)+([A-Za-z]+[\-\']?)*([A-Za-z]+)?$/

It expects optionally [at least 1 alphabetical character followed by a ' or -] an indefinite number of times. There must be at least one alphabetical character before a required space to ensure we are getting at least the first and last name. This entire pattern is grouped to accept indefinite repetition (for people who like to use all their names, such as John Jacob Jingleheimer Schmidt), but must appear at least once, by means of the + sign right in the middle. Finally, the last name is treated the same way as the other names, but no trailing space is allowed. (Unfortunately this means we are violating DRY a little bit.)

它期望可选 [至少 1 个字母字符后跟一个 ' 或 -] 无限次。在所需空格之前必须至少有一个字母字符，以确保我们至少获得名字和姓氏。整个模式被分组以接受无限重复（对于喜欢使用他们所有名字的人，例如 John Jacob Jingleheimer Schmidt），但必须至少出现一次，通过中间的 + 符号。最后，姓氏的处理方式与其他名称相同，但不允许尾随空格。（不幸的是，这意味着我们有点违反了 DRY。）

Here is the outcome on several possible pieces of input:

以下是几种可能的输入的结果：

"Jon Doe": true
"Jonathan Taylor Thomas": true
"Julia Louis-Dreyfus": true
"Jean-Paul Sartre": true
"Pat O'Brien": true
"Tór Eldon": false
"Marcus Wells-O'Shaugnessy": true
"Stephen Wells-O'Shaugnessy Marcus": true
"This-Is-A-Crazy-Name Jones": true
"---- --------": false
"'''' ''''''''": false
"'-'- -'-'-'-'": false
"a-'- b'-'-'-'": false
"'-'c -'-'-'-d": false
"e-'f g'-'-'-h": false
"'ij- -klmnop'": false

Note it still doesn't handle Unicode characters, but it could possibly be expanded to include those if needed.

请注意，它仍然不处理 Unicode 字符，但如果需要，它可能会扩展以包含这些字符。

回答by M.Georgiev

The below code works for my case:

以下代码适用于我的情况：

const isControlValid = control.value.match(/^[\x41-\x5A\x61-\x7A\xAA\xB5\xBA\xC0-\xD6\xD8-\xF6\xF8-\u02C1\u02C6-\u02D1\u02E0-\u02E4\u02EC\u02EE\u0370-\u0374\u0376\u0377\u037A-\u037D\u0386\u0388-\u038A\u038C\u038E-\u03A1\u03A3-\u03F5\u03F7-\u0481\u048A-\u0527\u0531-\u0556\u0559\u0561-\u0587\u05D0-\u05EA\u05F0-\u05F2\u0620-\u064A\u066E\u066F\u0671-\u06D3\u06D5\u06E5\u06E6\u06EE\u06EF\u06FA-\u06FC\u06FF\u0710\u0712-\u072F\u074D-\u07A5\u07B1\u07CA-\u07EA\u07F4\u07F5\u07FA\u0800-\u0815\u081A\u0824\u0828\u0840-\u0858\u08A0\u08A2-\u08AC\u0904-\u0939\u093D\u0950\u0958-\u0961\u0971-\u0977\u0979-\u097F\u0985-\u098C\u098F\u0990\u0993-\u09A8\u09AA-\u09B0\u09B2\u09B6-\u09B9\u09BD\u09CE\u09DC\u09DD\u09DF-\u09E1\u09F0\u09F1\u0A05-\u0A0A\u0A0F\u0A10\u0A13-\u0A28\u0A2A-\u0A30\u0A32\u0A33\u0A35\u0A36\u0A38\u0A39\u0A59-\u0A5C\u0A5E\u0A72-\u0A74\u0A85-\u0A8D\u0A8F-\u0A91\u0A93-\u0AA8\u0AAA-\u0AB0\u0AB2\u0AB3\u0AB5-\u0AB9\u0ABD\u0AD0\u0AE0\u0AE1\u0B05-\u0B0C\u0B0F\u0B10\u0B13-\u0B28\u0B2A-\u0B30\u0B32\u0B33\u0B35-\u0B39\u0B3D\u0B5C\u0B5D\u0B5F-\u0B61\u0B71\u0B83\u0B85-\u0B8A\u0B8E-\u0B90\u0B92-\u0B95\u0B99\u0B9A\u0B9C\u0B9E\u0B9F\u0BA3\u0BA4\u0BA8-\u0BAA\u0BAE-\u0BB9\u0BD0\u0C05-\u0C0C\u0C0E-\u0C10\u0C12-\u0C28\u0C2A-\u0C33\u0C35-\u0C39\u0C3D\u0C58\u0C59\u0C60\u0C61\u0C85-\u0C8C\u0C8E-\u0C90\u0C92-\u0CA8\u0CAA-\u0CB3\u0CB5-\u0CB9\u0CBD\u0CDE\u0CE0\u0CE1\u0CF1\u0CF2\u0D05-\u0D0C\u0D0E-\u0D10\u0D12-\u0D3A\u0D3D\u0D4E\u0D60\u0D61\u0D7A-\u0D7F\u0D85-\u0D96\u0D9A-\u0DB1\u0DB3-\u0DBB\u0DBD\u0DC0-\u0DC6\u0E01-\u0E30\u0E32\u0E33\u0E40-\u0E46\u0E81\u0E82\u0E84\u0E87\u0E88\u0E8A\u0E8D\u0E94-\u0E97\u0E99-\u0E9F\u0EA1-\u0EA3\u0EA5\u0EA7\u0EAA\u0EAB\u0EAD-\u0EB0\u0EB2\u0EB3\u0EBD\u0EC0-\u0EC4\u0EC6\u0EDC-\u0EDF\u0F00\u0F40-\u0F47\u0F49-\u0F6C\u0F88-\u0F8C\u1000-\u102A\u103F\u1050-\u1055\u105A-\u105D\u1061\u1065\u1066\u106E-\u1070\u1075-\u1081\u108E\u10A0-\u10C5\u10C7\u10CD\u10D0-\u10FA\u10FC-\u1248\u124A-\u124D\u1250-\u1256\u1258\u125A-\u125D\u1260-\u1288\u128A-\u128D\u1290-\u12B0\u12B2-\u12B5\u12B8-\u12BE\u12C0\u12C2-\u12C5\u12C8-\u12D6\u12D8-\u1310\u1312-\u1315\u1318-\u135A\u1380-\u138F\u13A0-\u13F4\u1401-\u166C\u166F-\u167F\u1681-\u169A\u16A0-\u16EA\u1700-\u170C\u170E-\u1711\u1720-\u1731\u1740-\u1751\u1760-\u176C\u176E-\u1770\u1780-\u17B3\u17D7\u17DC\u1820-\u1877\u1880-\u18A8\u18AA\u18B0-\u18F5\u1900-\u191C\u1950-\u196D\u1970-\u1974\u1980-\u19AB\u19C1-\u19C7\u1A00-\u1A16\u1A20-\u1A54\u1AA7\u1B05-\u1B33\u1B45-\u1B4B\u1B83-\u1BA0\u1BAE\u1BAF\u1BBA-\u1BE5\u1C00-\u1C23\u1C4D-\u1C4F\u1C5A-\u1C7D\u1CE9-\u1CEC\u1CEE-\u1CF1\u1CF5\u1CF6\u1D00-\u1DBF\u1E00-\u1F15\u1F18-\u1F1D\u1F20-\u1F45\u1F48-\u1F4D\u1F50-\u1F57\u1F59\u1F5B\u1F5D\u1F5F-\u1F7D\u1F80-\u1FB4\u1FB6-\u1FBC\u1FBE\u1FC2-\u1FC4\u1FC6-\u1FCC\u1FD0-\u1FD3\u1FD6-\u1FDB\u1FE0-\u1FEC\u1FF2-\u1FF4\u1FF6-\u1FFC\u2071\u207F\u2090-\u209C\u2102\u2107\u210A-\u2113\u2115\u2119-\u211D\u2124\u2126\u2128\u212A-\u212D\u212F-\u2139\u213C-\u213F\u2145-\u2149\u214E\u2183\u2184\u2C00-\u2C2E\u2C30-\u2C5E\u2C60-\u2CE4\u2CEB-\u2CEE\u2CF2\u2CF3\u2D00-\u2D25\u2D27\u2D2D\u2D30-\u2D67\u2D6F\u2D80-\u2D96\u2DA0-\u2DA6\u2DA8-\u2DAE\u2DB0-\u2DB6\u2DB8-\u2DBE\u2DC0-\u2DC6\u2DC8-\u2DCE\u2DD0-\u2DD6\u2DD8-\u2DDE\u2E2F\u3005\u3006\u3031-\u3035\u303B\u303C\u3041-\u3096\u309D-\u309F\u30A1-\u30FA\u30FC-\u30FF\u3105-\u312D\u3131-\u318E\u31A0-\u31BA\u31F0-\u31FF\u3400-\u4DB5\u4E00-\u9FCC\uA000-\uA48C\uA4D0-\uA4FD\uA500-\uA60C\uA610-\uA61F\uA62A\uA62B\uA640-\uA66E\uA67F-\uA697\uA6A0-\uA6E5\uA717-\uA71F\uA722-\uA788\uA78B-\uA78E\uA790-\uA793\uA7A0-\uA7AA\uA7F8-\uA801\uA803-\uA805\uA807-\uA80A\uA80C-\uA822\uA840-\uA873\uA882-\uA8B3\uA8F2-\uA8F7\uA8FB\uA90A-\uA925\uA930-\uA946\uA960-\uA97C\uA984-\uA9B2\uA9CF\uAA00-\uAA28\uAA40-\uAA42\uAA44-\uAA4B\uAA60-\uAA76\uAA7A\uAA80-\uAAAF\uAAB1\uAAB5\uAAB6\uAAB9-\uAABD\uAAC0\uAAC2\uAADB-\uAADD\uAAE0-\uAAEA\uAAF2-\uAAF4\uAB01-\uAB06\uAB09-\uAB0E\uAB11-\uAB16\uAB20-\uAB26\uAB28-\uAB2E\uABC0-\uABE2\uAC00-\uD7A3\uD7B0-\uD7C6\uD7CB-\uD7FB\uF900-\uFA6D\uFA70-\uFAD9\uFB00-\uFB06\uFB13-\uFB17\uFB1D\uFB1F-\uFB28\uFB2A-\uFB36\uFB38-\uFB3C\uFB3E\uFB40\uFB41\uFB43\uFB44\uFB46-\uFBB1\uFBD3-\uFD3D\uFD50-\uFD8F\uFD92-\uFDC7\uFDF0-\uFDFB\uFE70-\uFE74\uFE76-\uFEFC\uFF21-\uFF3A\uFF41-\uFF5A\uFF66-\uFFBE\uFFC2-\uFFC7\uFFCA-\uFFCF\uFFD2-\uFFD7\uFFDA-\uFFDC,'-.\s]+$/g);
const hasErrorCombinations = control.value.match(/(([^\x00-\x7F]+|[\w]+)?[.|\-|'|.|\s]{2,}([^\x00-\x7F]+|[\w]+)?)+|([^\x00-\x7F]+|[\w]+)?[.|\-|\s]{1}$|^[.|\-|\s]{1}([^\x00-\x7F]+|[\w]+)?$/g);
const hasAdditionalSymbolsCheckErrors = control.value.match(/[()+*]+/g);

return isControlValid && !hasErrorCombinations && !hasAdditionalSymbolsCheckErrors;

回答by unigg

Just add a space

加个空格就行

^[\w ]+$

Nor sure if it's your need.

也不确定这是否是您的需要。