【发布时间】:2018-03-19 15:15:31
【问题描述】:
例如,以下(无效)AU手机号码被libphonenumber ++++++614- -12a345678(())&认为是有效的:
代码方面:
final String mobilePhoneNumber = "++++++614- -12a345678(())&";
final String region = "AU";
final PhoneNumberUtil phoneNumberUtil = PhoneNumberUtil.getInstance();
// true
// I don't get it, how come ++++++614- -12a345678(())& is even considered a possible number??
System.out.println(phoneNumberUtil.isPossibleNumber(mobilePhoneNumber, region));
final Phonenumber.PhoneNumber phoneNumber = phoneNumberUtil.parse(mobilePhoneNumber, region);
// true
final boolean validNumberForRegion = phoneNumberUtil.isValidNumberForRegion(phoneNumber, region);
// true
final boolean validMobileNumber = phoneNumberUtil.getNumberType(phoneNumber).equals(PhoneNumberType.MOBILE);
在 libphonenumber v8.9.2 上测试:
compile group: 'com.googlecode.libphonenumber', name: 'libphonenumber', version: '8.9.2'
【问题讨论】:
标签: java validation phone-number libphonenumber