PHP正则表达式：每个单词必须以点结尾

Question

有人可以帮助我了解preg_match功能的具体模式吗？

• 字符串中的每个单词都必须以点
结尾
• 字符串的第一个字符必须是[a-zA-Z]
• 每个点后面都有一个空格
• 彼此不能有两个空格
• 最后一个字符必须是一个点（逐字逻辑）

示例：

• “Ing” - ＆gt;假
• “了Ing。” - ＆GT;真
• “ING。”。 - ＆GT;假
• “Xx Yy。” - ＆GT;假
• “XX。YY。” - ＆GT;真
• “XX.YY”。 - ＆GT;真

请问如何测试字符串？我的模式是

/^(([a-zA-Z]+)(?! ) \.)+\.$/

我知道这是错的，但我无法弄明白。感谢

Answer 1

检查这是否符合您的需求。

/^(?:[A-Z]+\. ?)+$/i

• ^匹配开始
• (?:打开non-capture group重复
• [A-Z]+与i flag匹配一个或多个alphas（较低和较高）
• \. ?匹配文字点后跟可选空格
• )+所有这一次或更长时间，直到$结束

Here's a demo at regex101

如果您想在结尾处禁用空格，请添加否定look：/^(?:[A-Z]+\. ?)+$(?<! )/i

Answer 2

试试这个：

$string = "Ing
Ing.
.Ing.
Xx Yy.
XX. YY.
XX.YY.";

if (preg_match('/^([A-Za-z]{1,}\.[ ]{0,})*/m', $string)) {
    // Successful match
} else {
    // Match attempt failed
}

结果：

正则表达式详细说明：

^               Assert position at the beginning of a line (at beginning of the string or after a line break character)
(               Match the regular expression below and capture its match into backreference number 1
   [A-Za-z]        Match a single character present in the list below
                      A character in the range between “A” and “Z”
                      A character in the range between “a” and “z”
      {1,}            Between one and unlimited times, as many times as possible, giving back as needed (greedy)
   \.              Match the character “.” literally
   [ ]             Match the character “ ”
      {0,}            Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
)*              Between zero and unlimited times, as many times as possible, giving back as needed (greedy)