Java Regex - 字母数字，允许前导空格但不允许空白字符串
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Java Regex - alphanumeric, allowing leading whitespace but not blank string
提问by deanmau5
I've been trying to make a java regex that allows only alphanumeric characters, which can have white spaces, BUT the whole string cannot be blank...
我一直在尝试制作一个只允许字母数字字符的 java 正则表达式，它可以有空格，但整个字符串不能为空...
Few examples..
几个例子..
" hello world123" //fine
"hello123world" //fine
"hello123world " //fine
" " //not allowed
So far I've gotten ^[a-zA-Z0-9][a-zA-Z0-9\s]*$though this does not allow any leading whitespace and so any string with x number leading whitespace is not being matched.
到目前为止，我已经得到 ^[a-zA-Z0-9][a-zA-Z0-9\s]*$虽然这不允许任何前导空格，因此任何带有 x 数字前导空格的字符串都不会匹配.
Any ideas what I could add to the expression to allow leading whitespace?
有什么想法可以添加到表达式中以允许前导空格吗？
回答by David says Reinstate Monica
How about just ^\s*[\da-zA-Z][\da-zA-Z\s]*$. 0 or more spaces at start, follow by at least 1 digit or letter followed by digits/letters/spaces.
刚刚怎么样^\s*[\da-zA-Z][\da-zA-Z\s]*$。开头有 0 个或多个空格，后跟至少 1 个数字或字母，后跟数字/字母/空格。
Note: I did not use \w because \w includes "_", which is not alphanumeric.
注意：我没有使用 \w，因为 \w 包括“_”，它不是字母数字。
Edit: Just tested all your cases on regexpal, and all worked as expected. This regex seems like the simplest one.
编辑：刚刚在 regexpal 上测试了所有案例，并且一切都按预期工作。这个正则表达式似乎是最简单的。
回答by Bohemian
Just use a look ahead to assert that there's at least one non-blank:
只需向前看即可断言至少有一个非空白：
(?=.*[^ ])[a-zA-Z0-9 ]+
This may be used as-is with String.matches():
这可以按原样使用String.matches()：
if (input.matches("(?=.*[^ ])[a-zA-Z0-9 ]+")) {
// input is OK
}
回答by Pshemo
You can test it using look-aheadmechanism ^(?=\\s*[a-zA-Z0-9])[a-zA-Z0-9\s]*$
您可以使用前瞻机制对其进行测试^(?=\\s*[a-zA-Z0-9])[a-zA-Z0-9\s]*$
^(?=\\s*[a-zA-Z0-9])will make regex to check if at start of the string it contains zero of more spaces \\s*and then character from class [a-zA-Z0-9].
^(?=\\s*[a-zA-Z0-9])将使正则表达式检查在字符串的开头它是否包含零个更多空格\\s*，然后是 class 中的字符[a-zA-Z0-9]。
Demo:
演示：
String[] data = {
" hello world123", //fine
"hello123world", //fine
"hello123world ", //fine
" " //not allowed
};
for(String s:data){
System.out.println(s.matches("(?=.*\S)[a-zA-Z0-9\s]*"));
}
output
输出
true
true
true
false
