java - 使用 Java 模式进行特殊字符检查

Question

我想用以下信息生成两种模式，

1) 账户表单中的名字、姓氏、电子邮件、电话号码字段中不能输入以下特殊字符：

模式 " [ ] : ; | = + * ? < > / \ ， 名称不能以句点开头

2）以下特殊字符不能在公司地址字段中输入：

模式< > / \ |

请给我一个想法。

提前致谢

Answer 1

试试这些模式

第 1 点

(?i)^([a-z][^"\[:\]\|=\+\*\?<>\\\/\r\n]+)$

第 2 点

(?i)^([a-z][^<>\\\/\|\r\n]+)$

解释

1st Pattern

"(?i)" +                               -- Match the remainder of the regex with the options: case insensitive (i)
"^" +                                  -- 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-z]" +                              -- Match a single character in the range between “a” and “z”
   "[^\"\\[:\\]\\|=\\+\\*\\?<>\\\\\\/\r\n]" +       -- Match a single character NOT present in the list below
                                             -- The character “"”
                                             -- A [ character
                                             -- The character “:”
                                             -- A ] character
                                             -- A | character
                                             -- The character “=”
                                             -- A + character
                                             -- A * character
                                             -- A ? character
                                             -- One of the characters “&lt;>”
                                             -- A \ character
                                             -- A / character
                                             -- A carriage return character
                                             -- A line feed character
      "+" +                                  -- Between one and unlimited times, as many times as possible, giving back as needed (greedy)
")" +
"$"                                    -- Assert position at the end of a line (at the end of the string or before a line break character)


2nd Pattern

"(?i)" +                  -- Match the remainder of the regex with the options: case insensitive (i)
"^" +                     -- 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-z]" +                 -- Match a single character in the range between “a” and “z”
   "[^<>\\\\\\/\\|\r\n]" +       -- Match a single character NOT present in the list below
                                -- One of the characters “&lt;>”
                                -- A \ character
                                -- A / character
                                -- A | character
                                -- A carriage return character
                                -- A line feed character
      "+" +                     -- Between one and unlimited times, as many times as possible, giving back as needed (greedy)
")" +
"$"                       -- Assert position at the end of a line (at the end of the string or before a line break character)

代码

try {
    boolean foundMatch = subjectString.matches("(?i)^([a-z][^\"\\[:\\]|=+*?<>\\\\/\\r\\n]+)$");
} catch (PatternSyntaxException ex) {
    // Syntax error in the regular expression
}

Answer 2

String.contains()您可以使用该方法，而不是使用您显然没有信心的正则表达式。

但是，如果您必须使用正则表达式，就像 Mayur Patel 所说的那样，“ [ab]”基本上意味着 a 或 b ！您应该查看正则表达式.info

Answer 3

以下是我的问题的解决方案

1) (?i)^([az][^\"\[:\]|=+*.?<>\\/\r\n]+)$

2) (?i)^([az][^\"<>|\\/\r\n]+)$

我还在 1) 点中添加了句点符号，用于检查不以句点符号开头的名称。

非常感谢 Cylian 和 Andy 的帮助，它真的帮了我很多。

再次感谢 ：）