javascript 正则表达式测试是否只有 ASCII 字符
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Regex to test if only ASCII characters
提问by Sami
I have tried this but it returns true for both UTF-8 and ASCII:
我试过这个，但它对于 UTF-8 和 ASCII 都返回 true：
console.log(/[^w/]+/.test("abc123")) //true
console.log(/[^w/]+/.test("???")) //true
回答by SmokeyPHP
I think you meant /[^\w]+/but what you really want, from what I can gather, is:
我想你的意思是，/[^\w]+/但你真正想要的是，据我所知，是：
console.log(/^[\x00-\x7F]+$/.test("abc123")) //true
console.log(/^[\x00-\x7F]+$/.test("abc_-8+")) //true
console.log(/^[\x00-\x7F]+$/.test("???")) //false
If you didn't actually mean to check the full ASCII set, you can just use:
如果您实际上并不是要检查完整的 ASCII 集，则可以使用：
console.log(/^[\w]+$/.test("abc123")) //true
console.log(/^[\w]+$/.test("abc_-8+")) //false
console.log(/^[\w]+$/.test("???")) //false
About \x notation
关于 \x 符号
\xFFis a hexadecimal notation (list here) used in this example for the range 00to 7Fto match the full ASCII character set. \x00-\x7Fis functionally indentical to a-zin that it specifies a range, however we are using hex notation for reliable ranging
\xFF是一个十六进制表示法（名单这里在本例中使用的范围内）00，以7F相匹配的完整的ASCII字符集。\x00-\x7F功能相同a-z，因为它指定了一个范围，但是我们使用十六进制表示法来实现可靠的范围
\wmatches 'word' characters, which is the same as [a-z0-9_]
\w匹配 'word' 字符，这与 [a-z0-9_]
