用逗号分隔的正则表达式电话号码

i need a regular expression to validate the field. It is a field of phone numbers, the number consists of 9 characters, numbers will be separated by commas.

我需要一个正则表达式来验证该字段。它是一个电话号码字段,数字由9个字符组成,数字将用逗号分隔。

Example:

123456789,123456789,123456789
123456789
123456789,123456789

I have written:

我已经写了:

~^(([0-9]{9,9},)+|([0-9]{9,9})+)$~i

but validating passes only numbers with commas.

但验证只传递带逗号的数字。

Example:

123456789,123456789,
123456789,

Can you help me with this?

你能帮我解决这个问题吗?

3 个解决方案

#1

You need to set a * quantifier to the first 9-digit group:

您需要将*量词设置为第一个9位数组:

~^(?:[0-9]{9},)*[0-9]{9}$~
               ^

See the regex demo. No need in ~i case insensitive modifier, there are no letters in the pattern. Also, {9,9} = {9}.

请参阅正则表达式演示。不需要~i不区分大小写的修饰符,模式中没有字母。此外,{9,9} = {9}。

The same pattern can be written as

相同的模式可以写成

~^[0-9]{9}(?:,[0-9]{9})*$~

See another demo.

看另一个演示。

Details:

^ - start of string

^ - 字符串的开头

(?:[0-9]{9},)* - 0+ sequences of:
- [0-9]{9} - 9 digits
- , - comma

(?:[0-9] {9},)* - 0+序列:[0-9] {9} - 9位数, - 逗号

[0-9]{9} - 9 digits

[0-9] {9} - 9位数

$ - end of string (may be replaced with \z to match the very end of string).

$ - 字符串结尾(可以用\ z替换以匹配字符串的结尾)。

EDIT: Since you only want to match comma-separated 9-digit chunks that only contain unique digits, it is possible to write a regex for that, although it is not recommended to use it. Best is to use your programming language means for that.

编辑:由于您只想匹配仅包含唯一数字的逗号分隔的9位数块,因此可以为其编写正则表达式,但不建议使用它。最好是使用您的编程语言手段。

A regex will look like (verbose version):

正则表达式看起来像(详细版本):

^                      # start of string
 (                     # Group 1 start
  (\d)                 # Digit 1 captured into Group 2
  (?!\2)(\d)           # Digit 2 not equal to the first one
  (?!\2|\3)(\d)        # etc.
  (?!\2|\3|\4)(\d)
  (?!\2|\3|\4|\5)(\d)
  (?!\2|\3|\4|\5|\6)(\d)
  (?!\2|\3|\4|\5|\6|\7)(\d)
  (?!\2|\3|\4|\5|\6|\7|\8)(\d)
  (?!\2|\3|\4|\5|\6|\7|\8|\9)(\d)
 )
 (?:,(?1))*        # 0+ sequences of , and the Group 1 pattern
$                  # End of string

See the regex demo. A one-liner:

请参阅正则表达式演示。单行:

^((\d)(?!\2)(\d)(?!\2|\3)(\d)(?!\2|\3|\4)(\d)(?!\2|\3|\4|\5)(\d)(?!\2|\3|\4|\5|\6)(\d)(?!\2|\3|\4|\5|\6|\7)(\d)(?!\2|\3|\4|\5|\6|\7|\8)(\d)(?!\2|\3|\4|\5|\6|\7|\8|\9)(\d))(?:,(?1))*$

In PHP, you may just use

在PHP中,您可以使用

if (count( array_unique( str_split( $s))) == strlen($s)) {
    echo "Unique";
} else {
    echo "Not unique";
}

#2

I don't think that your regex has to be very complicated:

我不认为你的正则表达式必须非常复杂:

^\d{9}(,\d{9})*,?

It will match at least one number block consisting of 9 digits.

它将匹配至少一个由9位数字组成的数字块。

Other 9 digit blocks might follow, if they are separated by a comma.

如果用逗号分隔,则可能会跟随其他9位数的块。

The whole expression can end with a comma.

整个表达式可以用逗号结尾。

#3

Just move that plus(+) to the end of the bigger group

只需将加号(+)移动到更大的组的末尾即可

^(([0-9]{9,9},)|([0-9]{9,9}))+$

#1