本文介绍了是否有电子邮件验证code Java ME的或BlackBerry?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
有没有Java ME的或BlackBerry一些标准的电子邮件验证code样?
解决方案
公共静态布尔validateEmailID(字符串email){
电子邮件= email.trim();
字符串反向=新的StringBuffer(电子邮件).reverse()的toString()。
如果(电子邮件== NULL || email.length()== 0 || email.indexOf(@)== -1){
返回false;
}
INT emailLength = email.length();
INT atPosition = email.indexOf(@);
INT atDot = reverse.indexOf(。); 串beforeAt = email.substring(0,atPosition);
串afterAt = email.substring(atPosition + 1,emailLength); 如果(beforeAt.length()== 0 || afterAt.length()== 0){
返回false;
}
的for(int i = 0; email.length() - 1 GT;我;我++){
炭的i1 = email.charAt(ⅰ);
炭I2 = email.charAt第(i + 1);
如果(I1 ==&放大器'。';&放大器; i2的=='。'){
返回false;
}
}
如果(email.charAt(atPosition - ''''''1)== || email.charAt(0)== || email.charAt(atPosition + 1)== || afterAt.indexOf( !@)= -1 || atDot 2){
返回false;
} 返回true;
}
Is there some standard email validator code sample for Java ME or BlackBerry?
解决方案
public static boolean validateEmailID(String email) {
email = email.trim();
String reverse = new StringBuffer(email).reverse().toString();
if (email == null || email.length() == 0 || email.indexOf("@") == -1) {
return false;
}
int emailLength = email.length();
int atPosition = email.indexOf("@");
int atDot = reverse.indexOf(".");
String beforeAt = email.substring(0, atPosition);
String afterAt = email.substring(atPosition + 1, emailLength);
if (beforeAt.length() == 0 || afterAt.length() == 0) {
return false;
}
for (int i = 0; email.length() - 1 > i; i++) {
char i1 = email.charAt(i);
char i2 = email.charAt(i + 1);
if (i1 == '.' && i2 == '.') {
return false;
}
}
if (email.charAt(atPosition - 1) == '.' || email.charAt(0) == '.' || email.charAt(atPosition + 1) == '.' || afterAt.indexOf("@") != -1 || atDot < 2) {
return false;
}
return true;
}
这篇关于是否有电子邮件验证code Java ME的或BlackBerry?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!