验证字符串以字母开头，至少包含3个字母，并且只包含字母、数字和下划线 - Validate string to start with a letter, contain at least 3 letters and contain only alphanumeric characters and underscores

Validate string to start with a letter, contain at least 3 letters and contain only alphanumeric characters and underscores

我正在尝试学习正则表达式，并试图为以下内容编写模式:

必须以字母开头
可以包含字母、数字或下划线。
必须在整个字符串中至少包含3个字母。

我已经试过了:

'#^[a-z0-9_]+$#i' // this one just matches the characters I want

这就是我卡住的地方:

'#^[a-z]{1}[a-z0-9_]+$#i' // doesn't work

我读了http://www.regular-expressions.info/reference.html，但还是卡住了

'#^[a-zA-Z]([0-9_]*[a-zA-Z]){2}[a-zA-Z0-9_]*$#'

[a-zA-Z]检查第一个约束(以字母开头)

([0-9_]*[a-zA-Z]){2}检查第二个约束(总共必须有3个字母)

[a-zA-Z0-9_]*释放了先前的限制，只限制使用允许的字符

应该这样做:

#^[a-zA-Z]{1}('w*[a-zA-Z]'w*){2}$#i

解释:

^[a-zA-Z]{1}  // 1st character is a letter
('w*[a-zA-Z]'w*){2}$ // The rest of the body contains exactly 2 letters, and 0 or more of any other word characters (letter, number, underscore)

/^(?:[a-z]'w*){3}$/i

此模式将要求完整字符串包含三个重复的字母，后面跟着零个或多个单词字符('w与[a-zA-Z0-9_]相同)。第三次重复中的'w将消耗字符串中所有剩余的单词字符。

代码(演示):

$tests = [
    '7hello',
    'Fooey',
    'Shoo3y',
    'Whoa!',
    'A1_B2_C3_',
    '_ABC',
    'UB40',
    'omg',
    'Hi_',
    'Stackoverflow_',
];
foreach ($tests as $test) {
    printf(
        "%s : %s'n",
        $test,
        preg_match('/^(?:[a-z]'w*){3}$/i', $test)
            ? 'pass'
            : 'fail'
    );
}

输出:

7hello : fail
Fooey : pass
Shoo3y : pass
Whoa! : fail
A1_B2_C3_ : pass
_ABC : fail
UB40 : fail
omg : pass
Hi_ : fail
Stackoverflow_ : pass