验证PAN卡号的正则表达式

我看过PAN103正常表达式的this question和this blog。 [A-Z]{5}[0-9]{4}[A-Z]{1}。但是我的问题比这个扩大了一点点。验证PAN卡号的正则表达式

在PAN卡号码：

1) The first three letters are sequence of alphabets from AAA to zzz 
2) The fourth character informs about the type of holder of the Card. Each assesse is unique:` 

    C — Company 
    P — Person 
    H — HUF(Hindu Undivided Family) 
    F — Firm 
    A — Association of Persons (AOP) 
    T — AOP (Trust) 
    B — Body of Individuals (BOI) 
    L — Local Authority 
    J — Artificial Judicial Person 
    G — Government 


3) The fifth character of the PAN is the first character 
    (a) of the surname/last name of the person, in the case of 
a "Personal" PAN card, where the fourth character is "P" or 
    (b) of the name of the Entity/ Trust/ Society/ Organisation 
in the case of Company/ HUF/ Firm/ AOP/ BOI/ Local Authority/ Artificial Jurdical Person/ Govt, 
where the fourth character is "C","H","F","A","T","B","L","J","G". 

4) The last character is a alphabetic check digit.

我想正则表达式要在此基础上检查。由于我得到了该人的姓名，或者在另一个EditText中获得了该组织名称，所以我需要进一步验证第4封和第5封信。

它原来是[A-Z]{3}[C,H,F,A,T,B,L,J,G,P]{1}**something for the fifth character**[0-9]{4}[A-Z]{1}

我无法弄清楚如何说东西已被写入。

以编程方式，它可以完成，someone has done it in rails但它可以通过正则表达式吗？怎么样？

来源

2015-05-27 inquisitive

做尝试（P | [C，H，F，A，T，B，L，J，G]） –

不能能够理解第三点。有多少个字符？ –

@AvinashRaj如果PAN卡号中的第四个字符是P，那么第五个字符将是您姓的第一个字母，例如。如果你是Avinash Raj，你的pancard数字的第4个字符将是P个人，第5个字符将是Raj的R.而如果PAN卡属于一个组织，比如说一家公司，那么第四个字符就是公司的C，第五个字母是公司名称的第一个字母，例如，如果有一家公司叫做“微软公司”，那么公司的第四个字符是“C”，第五个字符是微软的“M”。 – inquisitive

您可以使用matches()的正则表达式是基于用户的额外输入而形成的，并且look-behinds检查前面的第4个字符。如果第四个字母是P，我们为您在姓氏的第一个字母，如果第四个字母不是P，我们检查的实体名称的第一个字母：

String rx = "[A-Z]{3}([CHFATBLJGP])(?:(?<=P)" + c1 + "|(?<!P)" + c2 + ")[0-9]{4}[A-Z]";

Sample code：

String c1 = "S"; // First letter in surname coming from the EditText (with P before) 
String c2 = "F"; // First letter in name coming from another EditText (not with P before) 
String pan = "AWSPS1234Z"; // true 
System.out.println(pan.matches("[A-Z]{3}([CHFATBLJGP])(?:(?<=P)" + c1 + "|(?<!P)" + c2 + ")[0-9]{4}[A-Z]")); 
pan = "AWSCF1234Z"; // true 
System.out.println(pan.matches("[A-Z]{3}([CHFATBLJGP])(?:(?<=P)" + c1 + "|(?<!P)" + c2 + ")[0-9]{4}[A-Z]")); 
pan = "AWSCS1234Z"; // false 
System.out.println(pan.matches("[A-Z]{3}([CHFATBLJGP])(?:(?<=P)" + c1 + "|(?<!P)" + c2 + ")[0-9]{4}[A-Z]"));

来源

2015-05-27 07:43:51

Pan= edittextPan.getText().toString().trim(); 

Pattern pattern = Pattern.compile("[A-Z]{5}[0-9]{4}[A-Z]{1}"); 

Matcher matcher = pattern .matcher(Pan); 

if (matcher .matches()) { 
Toast.makeText(getApplicationContext(), Pan+" is Matching", 
Toast.LENGTH_LONG).show(); 

} 
else 
{ 
Toast.makeText(getApplicationContext(), Pan+" is Not Matching", 
Toast.LENGTH_LONG).show(); 
}

来源

2015-11-20 10:06:28

验证PAN卡号的正则表达式

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