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Java生成验证码

时间:2016-05-13 13:11:20      阅读:127      评论:0      收藏:0      [点我收藏+]

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                Random y=new Random();
		int s=y.nextInt(37);
		String a="s7q8r5s2b1g2d3s6a9f5c1v2f5h6j8g9p6h9f8c1";		
		System.out.println(a.substring(s, s+4));

基本思路:首先插入一段数字与字母的字符串;

              然后利用随机数生成随机数;

              然后用生成的随机数作为索引,截取该数的后面四位。

 

输出结果: 

技术分享 

 

 

第二种方法

 

基本思路

1、定义一个String[]zimu={"a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"};

2、生成随机数作为索引

3、生成10以内的数字

4、输出。

 

String[]zimu={"a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"};
	Random suoyin=new Random();
	Random shu=new Random();
	System.out.println(zimu[suoyin.nextInt(27)]+shu.nextInt(10)+zimu[suoyin.nextInt(27)]+shu.nextInt(10));

  

 

输出结果

 

技术分享

Java生成验证码

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原文地址:http://www.cnblogs.com/mutougezi/p/5487394.html

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