针对特定容易猜测的模式的 RegEx 验证 [英] RegEx Validation for Specific Easily Guessed Patterns

问题描述

所以我完全不知道从哪里开始在我的 React 应用程序中创建 RegEx 验证模式.

So I am completely not knowing in where to start in creating a RegEx validation pattern in my React application.

我有各种输入框，其中(取决于某些条件)会有 3、4、5 或 6 位密码(同样数量的输入框呈现在屏幕上，每个接受 1 个数字).

I have various input boxes, of which (depending on certain criteria) there will be either a 3, 4, 5, or 6 digit passcode (and that same number of input boxes rendered on screen to accept 1 number each).

s Save 或 Continue 按钮被点击后，单个输入被存储在一个数组中，然后joined 为一个数字.这是我必须验证的最终数字.

After s Save or Continue button is clicked, the individual inputs are stored in an array, and then joined to be one number. It is this final number that I must validate.

我需要做的是编写一个验证方法 - 在 React 中 - 它执行以下验证:

What I need to do is write a validation method - in React - which performs the following validations:

对于 3 位数字:

不能是 999、998 或紧跟在第一个数字之后的 11(911、611 等)

Cannot be 999, 998, or include 11 immediately after the first digit (911, 611, etc.)

对于 4 位数字:

不能是 9999、9998 或紧跟在第一个数字之后的 11(8112、5112 等)

Cannot be 9999, 9998, or include 11 immediately after the first digit (8112, 5112, etc.)

对于 5 位数字:

不能是 99999、99998 或紧跟在第一个数字之后的 11(71122、41122 等)

Cannot be 99999, 99998, or include 11 immediately after the first digit (71122, 41122, etc.)

对于 6 位数字:

不能是 999999、999998 或紧跟在第一个数字之后的 11(611222、311222 等)

Cannot be 999999, 999998, or include 11 immediately after the first digit (611222, 311222, etc.)

这样做让我很痛苦，但我确实缺乏 RegEx 空间，并且不确定如何实施这样的事情.

It pains me to do so, but I am really lacking in the RegEx space, and am uncertain how to go about implementing something like this.

还应该注意的是，我没有使用 Redux 或随之而来的验证.必须使用接受参数(从输入框检索到的最终数字)并针对该参数运行验证(如果失败则显示错误消息em>)

It should also be noted that I am not using Redux or the validations that come along with it. Gotta attack this one with a util method that accepts an argument (the final number retrieved from the input boxes) and runs the validation against that (displaying an error message if there is a failure)

有没有人可以提供一些有关如何开始使用 RegEx 的见解...?

Is there anyone that can provide some insight on how to go about getting started for the RegEx...?

提前致谢！

推荐答案

正如我在您的问题下方的评论，您可以使用以下正则表达式:

As in my comment below your question, you can use the following regex:

查看此处使用的正则表达式

^(?!9+[98]$|\d1{2})\d{3,6}$

工作原理:

• ^ 断言行首位置
• (?!9+[98]$|\d1{2}) 否定前瞻确保以下任一选项不会继续
  • 9+[98]$ 匹配 9 一次或多次，然后是 9 或 8，然后是行尾
  • \d1{2}) 匹配任何数字，后跟 1 两次
  • ^ assert position at the start of the line
  • (?!9+[98]$|\d1{2}) negative lookahead ensuring either of the the following options does not proceed
    • 9+[98]$ matches 9 one or more times, then either 9 or 8, then the end of the line
    • \d1{2}) matches any digit, followed by 1 twice

    由于负前瞻跟随行锚点的开始，我们也确保前瞻从那个位置开始，这就是为什么 \d1{2} 匹配 011, >111、211、...、911 而不是 1211 或其他.

    Since the negative lookahead follows the start of line anchor, we also ensure the lookahead starts at that position, that's why \d1{2} matches 011, 111, 211, ..., 911 and not 1211 or others.

    代码如下:

    s = ['999','998','911','611','9999','9998','8112','5112','99999','99998','71122','41122','999999','999998','611222','311222','123','6211','99989','121212']
    r = /^(?!9+[98]$|\d1{2})\d{3,6}$/
    for (x of s) {
      console.log(x.match(r) ? x + ': true' : x + ': false')
    }

    --

    OP 提到将 999 和 998 放在字符串中的任何位置应该使其无效:

    The OP mentioned that 999 and 998 placed anywhere in the string should invalidate it:

    查看此处使用的正则表达式

    ^(?!\d*9{2}[98]|\d1{2})\d{3,6}$
    

    与上面相同的正则表达式，除了负前瞻中的第一个选项.现在是 \d*9{2}[98]，匹配字符串中任意位置的 999 或 998(前面是任意数量的数字).

    Same regex as above except for the first option in the negative lookahead. It's now \d*9{2}[98], matching 999 or 998 anywhere in the string (preceded by any number of digits).

    s = ['999','998','911','611','9999','9998','8112','5112','99999','99998','71122','41122','999999','999998','611222','311222','123','6211','99989','121212']
    r = /^(?!\d*9{2}[98]|\d1{2})\d{3,6}$/
    for (x of s) {
      console.log(x.match(r) ? x + ': true' : x + ': false')
    }

    --

    OP 提到 0N11 的格式应该失效(不仅仅是 N11):

    The OP mentioned that the format of 0N11 should be invalidated (not just N11):

    查看此处使用的正则表达式

    ^(?!\d*9{2}[98]|[01]?\d1{2})\d{3,6}$
    

    与上面相同的正则表达式，除了负前瞻中的第二个选项.现在是 [01]?\d1{2}，可选匹配 0 或 1，后跟任意数字，然后 11 (so 011, 111, 211, ..., 911, 0011, 0111, 0211, 0311, ..., 0911, 1011、1111、1211、...、1911).

    Same regex as above except for the second option in the negative lookahead. It's now [01]?\d1{2}, matching 0 or 1 optionally, followed by any digit, then 11 (so 011, 111, 211, ..., 911, 0011, 0111, 0211, 0311, ..., 0911, 1011, 1111, 1211, ..., 1911).

    s = ['999','998','911','611','9999','9998','8112','5112','99999','99998','71122','41122','999999','999998','611222','311222','123','6211','99989','121212']
    r = /^(?!\d*9{2}[98]|[01]?\d1{2})\d{3,6}$/
    for (x of s) {
      console.log(x.match(r) ? x + ': true' : x + ': false')
    }

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