javascript 正则表达式：只有字母数字但不是纯数字
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Regex: only alphanumeric but not if this is pure numeric
提问by oldergod
For instance:
例如：
'1' => NG
'243' => NG
'1av' => OK
'pRo' => OK
'123k%' => NG
I tried with
我试过
/^(?=^[^0-9]*$)(?=[a-zA-Z0-9]+)$/
but it is not working very well.
但它工作得不是很好。
回答by Jens
回答by Andrzej Doyle
So we know that there must be at least one "alphabetic" character in there somewhere:
所以我们知道某处必须至少有一个“字母”字符：
[a-zA-Z]
And it can have any number of alphanumeric characters (including zero) either before it or after it, so we pad it with [a-zA-Z0-9]*on both sides:
它可以在它之前或之后有任意数量的字母数字字符（包括零），所以我们在它的[a-zA-Z0-9]*两边填充它：
/^[a-zA-Z0-9]*[a-zA-Z][a-zA-Z0-9]*$/
That should do the trick.
这应该够了吧。
回答by MaxArt
Try with this:
试试这个：
/^(?!^\d*$)[a-zA-Z\d]*$/
Edit: since this is essentially the same of the accepted answer, I'll give you something different:
编辑：由于这与接受的答案基本相同，我会给你一些不同的东西：
/^\d*[a-zA-Z][a-zA-Z\d]*$/
No lookaheads, no repeated complex group, it just verifies that there's at least one alphabetic character. This should be quite fast.
没有前瞻，没有重复的复杂组，它只是验证至少有一个字母字符。这应该是相当快的。
回答by godspeedlee
try with this: ^(?!\d+\b)[a-zA-Z\d]+$(?!\d+\b)will avoid pure numeric, add \w+ then mix alphabet and number.
试试这个：^(?!\d+\b)[a-zA-Z\d]+$(?!\d+\b)将避免纯数字，添加 \w+ 然后混合字母和数字。
回答by user2184688
Try this: ^[a-zA-Z0-9]*[a-zA-Z]+[a-zA-Z0-9]*$
试试这个： ^[a-zA-Z0-9]*[a-zA-Z]+[a-zA-Z0-9]*$
回答by Abhishek Gehlot
this is strictly alphanumeric
这是严格的字母数字
String name="^[^a-zA-Z]\d*[a-zA-Z][a-zA-Z\d]*$";
回答by davidrac
This should do the trick, without lookbehind:
这应该可以解决问题，无需回头看：
^(\d*[a-zA-Z]\d*)+$
