public static void main(String[] args){
Test t = new Test()
String email = "[email protected]"
boolean isEmail = t.isEmail(email)
if(isEmail){
System.out.println(email +" is a right email!!!")
}else{
System.out.println("error")
}
}
public boolean isEmail(String email){
//正则表达式
/*
String regex = "^[A-Za-z]{1,40}@[A-Za-z0-9]{1,40}\\.[A-Za-z]{2,3}$"
return email.matches(regex)
*/
//不适用正则
if(email==null||"".equals(email)) return false
if(!containsOneWord('@',email)||!containsOneWord('.',email)) return false
String prefix = email.substring(0,email.indexOf("@"))
String middle = email.substring(email.indexOf("@")+1,email.indexOf("."))
String subfix = email.substring(email.indexOf(".")+1)
System.out.println("prefix="+prefix +" middle="+middle+" subfix="+subfix)
if(prefix==null||prefix.length()>40||prefix.length()==0) return false
if(!isAllWords(prefix)) return false
if(middle==null||middle.length()>40||middle.length()==0) return false
if(!isAllWordsAndNo(middle)) return false
if(subfix==null||subfix.length()>3||subfix.length()<2) return false
if(!isAllWords(subfix)) return false
return true
}
//判断字符串只包含指定的一个字符c
private boolean containsOneWord(char c , String word){
char[] array = word.toCharArray()
int count = 0
for(Character ch : array){
if(c == ch) {
count++
}
}
return count==1
}
//检查一个字符串是否全部是字母
private boolean isAllWords(String prefix){
char[] array = prefix.toCharArray()
for(Character ch : array){
if(ch<'A' || ch>'z' || (ch<'a' &&ch>'Z')) return false
}
return true
}
//检查一个字符串是否包含字母和数字
private boolean isAllWordsAndNo(String middle){
char[] array = middle.toCharArray()
for(Character ch : array){
if(ch<'0' || ch >'z') return false
else if(ch >'9' &&ch <'A') return false
else if(ch >'Z' &&ch <'a') return false
}
return true
}
}
自己写逻辑验证的方式可以做到。我写了下面的：public static boolean validateEmail(String email) {
boolean flag = false
int pos = email.indexOf("@")
if (pos == -1 || pos == 0 || pos == email.length() - 1) {
return false
}
String[] strings = email.split("@")
if (strings.length != 2) {// 如果邮箱不是xxx@xxx格式
return false
}
CharSequence cs = strings[0]
for (int i = 0i <cs.length()i++) {
char c = cs.charAt(i)
if (!Character.isLetter(c) &&!Character.isDigit(c)) {
return false
}
}
pos = strings[1].indexOf(".")// 如果@后面没有.，则是错误的邮箱。
if (pos == -1 || pos == 0 || pos == email.length() - 1) {
return false
}
strings = strings[1].split(".")
for (int j = 0j <strings.lengthj++) {
cs = strings[j]
if (cs.length() == 0) {
return false
}
for (int i = 0i <cs.length()i++) {//如果保护不规则的字符，表示错误
char c = cs.charAt(i)
if (!Character.isLetter(c) &&!Character.isDigit(c)) {
return false
}
}
}
return true
}
这个效率也不会差很多，不过我推荐matches方法，经过测试的，matches匹配方式运行10000次的时间比上面的时间少了20毫秒。
