java中的电话号码验证
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phone Number validation in java
提问by Devendra
I want to validate a phone number in such Way :-
我想以这种方式验证电话号码：-
The field should allow the user to enter characters and should auto-correct. So an entry of "+1-908-528-5656" would not create an error for the user, it would just change to "19085285656".
该字段应允许用户输入字符并应自动更正。因此，输入“+1-908-528-5656”不会为用户创建错误，它只会更改为“19085285656”。
I also want to number range between 9 to 11.
我也想编号范围在 9 到 11 之间。
I also tried with the below code but not concluded to the final solution:
我也尝试了下面的代码，但没有得出最终的解决方案：
final String PHONE_REGEX = "^\+([0-9\-]?){9,11}[0-9]$";
final Pattern pattern = Pattern.compile(PHONE_REGEX);
String phone = "+1-908-528-5656";
phone=phone.replaceAll("[\-\+]", "");
System.out.println(phone);
final Matcher matcher = pattern.matcher(phone);
System.out.println(matcher.matches());
采纳答案by Braj
You can use simple String.matches(regex)to test any string against a regex pattern instead of using Patternand Matcherclasses.
您可以使用 simpleString.matches(regex)来针对正则表达式模式测试任何字符串，而不是使用Pattern和Matcher类。
Sample:
样本：
boolean isValid = phoneString.matches(regexPattern);
Here is the regex pattern as per your input string:
这是根据您的输入字符串的正则表达式模式：
\+\d(-\d{3}){2}-\d{4}
Better use Spring validation annotationfor validation.
最好使用Spring 验证注解进行验证。
回答by k_g
Assuming you want an optimization (which is what your comment suggests).
假设您想要优化（这就是您的评论所建议的）。
How bout this? (the "0" is to exclude if they give complete garbage without even a single digit).
这个怎么样？（“0”是排除如果他们给出完整的垃圾，甚至没有一个数字）。
int parse(String phone){
int num = Integer.parseInt("0"+phone.replaceAll("[^0-9]",""));
return 100000000<=num&&num<100000000000?num:-1;
}
回答by Md. kamrul Hasan
Assuming your input field take any kind of character and you just want the digits.
假设您的输入字段采用任何类型的字符，而您只需要数字。
String phone = "+1-908-528-5656";
phone=phone.replaceAll("[\D]","");
if(phone.length()>=9 || phone.length()<=11)
System.out.println(phone);
回答by Deepak
I am not sure but removing the garbage characters parenthesis, spaces and hyphens, if you match with ^((\+[1-9]?[0-9])|0)?[7-9][0-9]{9}$ , you may validate a mobile number
我不确定但删除垃圾字符括号、空格和连字符，如果你匹配^((\+[1-9]?[0-9])|0)?[7-9][0-9] {9}$，您可以验证手机号码
private static final String PHONE_NUMBER_GARBAGE_REGEX = "[()\s-]+";
private static final String PHONE_NUMBER_REGEX = "^((\+[1-9]?[0-9])|0)?[7-9][0-9]{9}$";
private static final Pattern PHONE_NUMBER_PATTERN = Pattern.compile(PHONE_NUMBER_REGEX);
public static boolean validatePhoneNumber(String phoneNumber) {
return phoneNumber != null && PHONE_NUMBER_PATTERN.matcher(phoneNumber.replaceAll(PHONE_NUMBER_GARBAGE_REGEX, "")).matches();
}
回答by rohan
We can use String.matches(String regex)1to validate phone numbers using java.
我们可以使用String.matches(String regex)1来验证使用 java 的电话号码。
Sample code snippet
示例代码片段
package regex;
public class Phone {
private static boolean isValid(String s) {
String regex = "\d{3}-\d{3}-\d{4}"; // XXX-XXX-XXXX
return s.matches(regex);
}
public static void main(String[] args) {
System.out.println(isValid("123-456-7890"));
}
}
P.S. The regex pattern we use extra '\' for escaping when we use in java string. (Try to use "\d{3}-\d{3}-\d{4}" in java program, you will get an error.
PS 当我们在 java 字符串中使用时，我们使用额外的 '\' 进行转义的正则表达式模式。（在java程序中尝试使用“\d{3}-\d{3}-\d{4}”，会报错。
回答by jessica
// The Regex not validate mobile number, which is in internation format.
// The Following code work for me.
// I have use libphonenumber library to validate Number from below link.
// http://repo1.maven.org/maven2/com/googlecode/libphonenumber/libphonenumber/8.0.1/
// https://github.com/googlei18n/libphonenumber
// Here, is my source code.
public boolean isMobileNumberValid(String phoneNumber)
{
boolean isValid = false;
// Use the libphonenumber library to validate Number
PhoneNumberUtil phoneUtil = PhoneNumberUtil.getInstance();
Phonenumber.PhoneNumber swissNumberProto =null ;
try {
swissNumberProto = phoneUtil.parse(phoneNumber, "CH");
} catch (NumberParseException e) {
System.err.println("NumberParseException was thrown: " + e.toString());
}
if(phoneUtil.isValidNumber(swissNumberProto))
{
isValid = true;
}
// The Library failed to validate number if it contains - sign
// thus use regex to validate Mobile Number.
String regex = "[0-9*#+() -]*";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(phoneNumber);
if (matcher.matches()) {
isValid = true;
}
return isValid;
}
回答by Armer B.
One easy and simple to use java phone validation regex:
一个简单易用的 java 手机验证正则表达式：
public static final String PHONE_VERIFICATION = "^[+0-9-\(\)\s]*{6,14}$";
private static Pattern p;
private static Matcher m;
public static void main(String[] args)
{
//Phone validation
p = Pattern.compile(PHONE_VERIFICATION);
m = p.matcher("+1 212-788-8609");
boolean isPhoneValid = m.matches();
if(!isPhoneValid)
{
System.out.println("The Phone number is NOT valid!");
return;
}
System.out.println("The Phone number is valid!");
}
回答by Deepesh Jain
i have done testing one regex for this combination of phone numbers
我已经为这种电话号码组合测试了一个正则表达式
(294) 784-4554
(247) 784 4554
(124)-784 4783
(124)-784-4783
(124) 784-4783
+1(202)555-0138
THIS REGEX SURELY WILL BE WORKING FOR ALL THE US NUMBERS
这个正则表达式肯定适用于所有美国号码
\d{10}|(?:\d{3}-){2}\d{4}|\(\d{3}\)\d{3}-?\d{4}|\(\d{3}\)-\d{3}-?\d{4}|\(\d{3}\) \d{3} ?\d{4}|\(\d{3}\)-\d{3} ?\d{4}|\(\d{3}\) \d{3}-?\d{4}
