mysql判断邮件地址是不是合法

Posted 2023-04-22

参考技术A import java.io.*;
import java.net.*;
import java.util.*;
import javax.naming.*;
import javax.naming.directory.*;

public class SMTPMXLookup
private static int hear( BufferedReader in ) throws IOException
String line = null;
int res = 0;

while ( (line = in.readLine()) != null )
String pfx = line.substring( 0, 3 );
try
res = Integer.parseInt( pfx );

catch (Exception ex)
res = -1;

if ( line.charAt( 3 ) != '-' ) break;


return res;


private static void say( BufferedWriter wr, String text )
throws IOException
wr.write( text + "\r\n" );
wr.flush();

return;

private static ArrayList getMX( String hostName )
throws NamingException
// Perform a DNS lookup for MX records in the domain
Hashtable env = new Hashtable();
env.put("java.naming.factory.initial",
"com.sun.jndi.dns.DnsContextFactory");
DirContext ictx = new InitialDirContext( env );
Attributes attrs = ictx.getAttributes
( hostName, new String[] "MX" );
Attribute attr = attrs.get( "MX" );

// if we don't have an MX record, try the machine itself
if (( attr == null ) || ( attr.size() == 0 ))
attrs = ictx.getAttributes( hostName, new String[] "A" );
attr = attrs.get( "A" );
if( attr == null )
throw new NamingException
( "No match for name '" + hostName + "'" );

// Huzzah! we have machines to try. Return them as an array list
// NOTE: We SHOULD take the preference into account to be absolutely
// correct. This is left as an exercise for anyone who cares.
ArrayList res = new ArrayList();
NamingEnumeration en = attr.getAll();

while ( en.hasMore() )
String mailhost;
String x = (String) en.next();
String f[] = x.split( " " );
// THE fix *************
if (f.length == 1)
mailhost = f[0];
else if ( f[1].endsWith( "." ) )
mailhost = f[1].substring( 0, (f[1].length() - 1));
else
mailhost = f[1];
// THE fix *************
res.add( mailhost );

return res;


public static boolean isAddressValid( String address )
// Find the separator for the domain name
int pos = address.indexOf( '@' );

// If the address does not contain an '@', it's not valid
if ( pos == -1 ) return false;

// Isolate the domain/machine name and get a list of mail exchangers
String domain = address.substring( ++pos );
ArrayList mxList = null;
try
mxList = getMX( domain );

catch (NamingException ex)
return false;


// Just because we can send mail to the domain, doesn't mean that the
// address is valid, but if we can't, it's a sure sign that it isn't
if ( mxList.size() == 0 ) return false;

// Now, do the SMTP validation, try each mail exchanger until we get
// a positive acceptance. It *MAY* be possible for one MX to allow
// a message [store and forwarder for example] and another [like
// the actual mail server] to reject it. This is why we REALLY ought
// to take the preference into account.
for ( int mx = 0 ; mx < mxList.size() ; mx++ )
boolean valid = false;
try
int res;
//
Socket skt = new Socket( (String) mxList.get( mx ), 25 );
BufferedReader rdr = new BufferedReader
( new InputStreamReader( skt.getInputStream() ) );
BufferedWriter wtr = new BufferedWriter
( new OutputStreamWriter( skt.getOutputStream() ) );

res = hear( rdr );
if ( res != 220 ) throw new Exception( "Invalid header" );
say( wtr, "EHLO rgagnon.com" );

res = hear( rdr );
if ( res != 250 ) throw new Exception( "Not ESMTP" );

// validate the sender address
say( wtr, "MAIL FROM: <[e-mail]>" );
res = hear( rdr );
if ( res != 250 ) throw new Exception( "Sender rejected" );

say( wtr, "RCPT TO: <" + address + ">" );
res = hear( rdr );

// be polite
say( wtr, "RSET" ); hear( rdr );
say( wtr, "QUIT" ); hear( rdr );
if ( res != 250 )
throw new Exception( "Address is not valid!" );

valid = true;
rdr.close();
wtr.close();
skt.close();

catch (Exception ex)
// Do nothing but try next host
ex.printStackTrace();

finally
if ( valid ) return true;


return false;


public static void main( String args[] )
String testData[] =
"[e-mail]",
"[e-mail]",
"[e-mail]", // Invalid domain name
"[e-mail]", // Invalid address
"[e-mail]" // Failure of this method
;

for ( int ctr = 0 ; ctr < testData.length ; ctr++ )
System.out.println( testData[ ctr ] + " is valid? " +
isAddressValid( testData[ ctr ] ) );

return;

算法判断IP地址是不是合法的，包含IPv4和IPv6

1. 概述

类似题目：

LeetCode468: https://leetcode.com/problems/validate-ip-address/

LeetCode93: https://leetcode.com/problems/restore-ip-addresses/

编写一个函数来验证输入的每个字符串是否是有效的IPv4或IPv6地址。

• 如果是有效的 IPv4 地址，返回 “IPv4”；
• 如果是有效的 IPv6 地址，发挥 “IPv6”；
• 如果不是上述类型的IP地址，返回 “Neither”。

IPv4地址由十进制数和点来表示，每个地址包含4个十进制数，其范围为 0 ~ 255，用 (“.”) 分割，比如：
172.16.254.1

同时，IPv4地址内的数不会以0开头。比如，地址 172.16.254.1 是不合法的。

IPv6地址由8组16进制的数字来表示，每组表示16比特。这些数字通过 (“:”) 分割，比如：2001:0db8:85a3:0000:0000:8a2e:0370:7334 是一个有效的地址。而且，我们可以加入一些以0开头的数字，字母可以使用大写，也可以是小写。所以，2001:0db8:85a3:0000:0000:8A2E:0370:7334 也是一个有效的 IPv6 地址（即，忽略0开头，忽略大小写）。

然而，我们不能因为某个组的值为0，而使用一个空的组，以至于出现（::）的情况。比如，2001:0db8:85a3::8A2E:0370:7334 是无效的 IPv6 地址。

同时，在IPv6地址中，多余的0也是不被允许的。比如，02001:0db8:85a3:0000:0000:8a2e:0370:7334 是无效的。

例子

输入：IP = “172.16.254.1”
输出：“IPv4”
解释：有效的 IPv4 地址，返回 “IPv4”

2. 代码及测试

Java

方法1

class Solution 
    public String validIPAddress(String queryIP) 
        if (queryIP == null || queryIP.length() < 1) return "Neither";
        int n = queryIP.length();
        int index = 0;
        while (index < n && queryIP.charAt(index) != '.' && queryIP.charAt(index) != ':') index++;
        if (index == n) return "Neither";
        int cnt = 0;
        if (queryIP.charAt(index) == '.') 
            // check IPv4
            index = 0;
            int left = 0;
            while (index < n) 
                while (index < n && queryIP.charAt(index) != '.') index++;
                String str = queryIP.substring(left, index);
                if (cnt > 4 || str == null || str.length() < 1 || str.length() > 3 || (str.length() > 1 && str.charAt(0) == '0')) return "Neither";
                for (int i = 0; i < str.length(); i++) 
                    if (str.charAt(i) < '0' || str.charAt(i) > '9') return "Neither";
                
                int val = Integer.parseInt(str);
                if (val < 0 || val > 255) return "Neither";
                cnt++;
                left = index + 1;
                index++;
            
            if (cnt == 4 && index == n + 1) return "IPv4";
         else 
            // check IPv6
            index = 0;
            int left = 0;
            while (index < n) 
                while (index < n && queryIP.charAt(index) != ':') index++;
                String str = queryIP.substring(left, index);
                if (cnt > 8 || str == null || str.length() < 1 || str.length() > 4) return "Neither";
                for (int i = 0; i < str.length(); i++) 
                    if (!(str.charAt(i) >= '0' && str.charAt(i) <= '9') && !(str.charAt(i) >= 'a' && str.charAt(i) <= 'f') && !(str.charAt(i) >= 'A' && str.charAt(i) <= 'F')) return "Neither";
                
                cnt++;
                left = index + 1;
                index++;
            
            if (cnt == 8 && index == n + 1) return "IPv6";
        
        return "Neither";
    

方法2

class Solution 
    public String validIPAddress(String queryIP) 
        if (queryIP == null || queryIP.length() < 1) 
            return "Neither";
        
        queryIP = queryIP.toLowerCase();
        int n = queryIP.length();
        if (queryIP.charAt(n - 1) == '.' || queryIP.charAt(n - 1) == ':') 
            return "Neither";
        
        int index = 0;
        while (index < n && queryIP.charAt(index) != '.' && queryIP.charAt(index) != ':') index++;
        if (index == n) return "Neither";
        if (queryIP.charAt(index) == '.') 
            // check IPv4
            String[] strs = queryIP.split("\\\\.");
            if (strs.length != 4) return "Neither";
            for (int i = 0; i < 4; i++) 
                if (strs[i].length() < 1 || strs[i].length() > 3 || (strs[i].length() > 1 && strs[i].charAt(0) == '0')) return "Neither";
                for (int j = 0; j < strs[i].length(); j++) 
                    if (strs[i].charAt(j) < '0' || strs[i].charAt(j) > '9') return "Neither";
                
                int val = Integer.parseInt(strs[i]);
                if (val < 0 || val > 255) return "Neither";
            
            return "IPv4";
         else 
            // check IPv6
            String[] strs = queryIP.split(":");
            if (strs.length != 8) return "Neither";
            for (int i = 0; i < 8; i++) 
                if (strs[i].length() < 1 || strs[i].length() > 4) return "Neither";
                for (int j = 0; j < strs[i].length(); j++) 
                    if (!(strs[i].charAt(j) >= '0' && strs[i].charAt(j) <= '9') && !(strs[i].charAt(j) >= 'a' && strs[i].charAt(j) <= 'f')) 
                        return "Neither";
                    
                
            
            return "IPv6";
        
    

C++

方法1

class Solution 
public:
    string validIPAddress(string queryIP) 
        int n = queryIP.size();
        if(n < 7) return "Neither";
        int index = 0;
        while (index < n && queryIP[index] != '.' && queryIP[index] != ':') index++;
        if (index == n) return "Neither";
        int cnt = 0;
        if (queryIP[index] == '.') 
            // check IPv4
            index = 0;
            int left = index;
            while (index < n) 
                while (index < n && queryIP[index] != '.') index++;
                string t = queryIP.substr(left, index - left);
                if (cnt > 4 || t.empty() || (t.size() > 1 && t[0] == '0') || t.size() > 3) return "Neither";
                for (char c: t) 
                    if (c < '0' || c > '9') return "Neither";
                
                int val = stoi(t);
                if (val < 0 || val > 255) return "Neither";
                cnt++;
                left = index + 1;
                index++;
            
            if (cnt == 4 && index == n + 1) return "IPv4";
         else 
            // check IPv6
            index = 0;
            int left = index;
            while (index < n) 
                while (index < n && queryIP[index] != ':') index++;
                string t = queryIP.substr(left, index - left);
                if (cnt > 8 || t.empty() || t.size() > 4) return "Neither";
                for (char c: t) 
                    if (!(c >= '0' && c <= '9') && !(c >= 'a' && c <= 'f') && !(c >= 'A' && c <= 'F')) return "Neither";
                
                cnt++;
                left = index + 1;
                index++;
            
            if (cnt == 8 && index == n + 1) return "IPv6";
        
        return "Neither";
    
;

方法2

class Solution 
public:
    string validIPAddress(string IP) 
        istringstream is(IP);
        string t = "";
        int cnt = 0;
        if (IP.find(':') == string::npos)  // Check IPv4
            while (getline(is, t, '.')) 
                ++cnt;
                if (cnt > 4 || t.empty() || (t.size() > 1 && t[0] == '0') || t.size() > 3) return "Neither";
                for (char c: t) 
                    if (c < '0' || c > '9') return "Neither";
                
                int val = stoi(t);
                if (val < 0 || val > 255) return "Neither";
            
            return (cnt == 4 && IP.back() != '.') ? "IPv4" : "Neither";
         else  // Check IPv6
            while (getline(is, t, ':')) 
                ++cnt;
                if (cnt > 8 || t.empty() || t.size() > 4) return "Neither";
                for (char c: t) 
                    if (!(c >= '0' && c <= '9') && !(c >= 'a' && c <= 'f') && !(c >= 'A' && c <= 'F')) return "Neither";
                
            
            return (cnt == 8 && IP.back() != ':') ? "IPv6" : "Neither";
        
    
;