vb.net 是否可以验证 IMEI 号码？
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Is it possible to validate IMEI Number?
提问by Kichu
For a mobile shop application, I need to validate an IMEI number. I know how to validate based on input length, but is their any other mechanism for validating the input number? Is there any built-in function that can achieve this?
对于移动商店应用程序，我需要验证IMEI 号码。我知道如何根据输入长度进行验证，但是它们是否还有其他用于验证输入数字的机制？是否有任何内置功能可以实现这一点？
Logic from any language is accepted, and appreciated.
任何语言的逻辑都被接受和赞赏。
回答by Karl Nicoll
A search suggests that there isn't a built-infunction that will validate an IMEI number, but there is a validation method using the Luhn algorithm.
搜索表明没有内置函数可以验证 IMEI 号码，但有一种使用Luhn 算法的验证方法。
General process:
一般流程：
- Input IMEI:
490154203237518 - Take off the last digit, and remember it:
49015420323751&8. This last digit 8 is the validation digit. - Double each second digit in the IMEI:
4 18 0 2 5 8 2 0 3 4 3 14 5 2(excluding the validation digit) - Separate this number into single digits:
4 1 8 0 2 5 8 2 0 3 4 3 1 4 5 2(notice that18and14have been split). - Add up all the numbers:
4+1+8+0+2+5+8+2+0+3+4+3+1+4+5+2=52 - Take your resulting number, remember it, and round it up to the nearest multiple of ten:
60. - Subtract your original number from the rounded-up number:
60 - 52=8. - Compare the result to your original validation digit. If the two numbers match, your IMEI is valid.
- 输入IMEI：
490154203237518 - 去掉最后一个数字，记住它：
49015420323751&8。最后一位数字 8 是验证数字。 - IMEI 中每第二个数字加倍：（
4 18 0 2 5 8 2 0 3 4 3 14 5 2不包括验证数字） - 将此数字分隔为单个数字：（
4 1 8 0 2 5 8 2 0 3 4 3 1 4 5 2注意18和14已被拆分）。 - 把所有数字加起来：
4+1+8+0+2+5+8+2+0+3+4+3+1+4+5+2=52 - 取你得到的数字，记住它，然后把它四舍五入到最接近的十的倍数：
60。 - 从四舍五入的数字中减去您的原始数字：
60 - 52=8。 - 将结果与您的原始验证数字进行比较。如果这两个数字匹配，则您的 IMEI 有效。
The IMEI given in step 1 above is valid, because the number found in step #7 is 8, which matches the validation digit.
上面第 1 步中给出的 IMEI 是有效的，因为在第 7 步中找到的数字是 8，它与验证数字匹配。
回答by Osvel Alvarez Jacomino
According to the previous answerfrom Karl Nicolli'm created this method in Java.
根据Karl Nicoll之前的回答，我在 Java 中创建了这个方法。
public static int validateImei(String imei) {
//si la longitud del imei es distinta de 15 es invalido
if (imei.length() != 15)
return CheckImei.SHORT_IMEI;
//si el imei contiene letras es invalido
if (!PhoneNumber.allNumbers(imei))
return CheckImei.MALFORMED_IMEI;
//obtener el ultimo digito como numero
int last = imei.charAt(14) - 48;
//duplicar cada segundo digito
//sumar cada uno de los digitos resultantes del nuevo imei
int curr;
int sum = 0;
for (int i = 0; i < 14; i++) {
curr = imei.charAt(i) - 48;
if (i % 2 != 0){
// sum += duplicateAndSum(curr);
// initial code from Osvel Alvarez Jacomino contains 'duplicateAndSum' method.
// replacing it with the implementation down here:
curr = 2 * curr;
if(curr > 9) {
curr = (curr / 10) + (curr - 10);
}
sum += curr;
}
else {
sum += curr;
}
}
//redondear al multiplo de 10 superior mas cercano
int round = sum % 10 == 0 ? sum : ((sum / 10 + 1) * 10);
return (round - sum == last) ? CheckImei.VALID_IMEI_NO_NETWORK : CheckImei.INVALID_IMEI;
}
回答by Istvan Darvas
IMEI can start with 0 digit. This is why the function input is string. Thanks for the method @KarlNicol
IMEI 可以从 0 位数字开始。这就是函数输入是字符串的原因。感谢@KarlNicol的方法
Golang
高朗
func IsValid(imei string) bool {
digits := strings.Split(imei, "")
numOfDigits := len(digits)
if numOfDigits != 15 {
return false
}
checkingDigit, err := strconv.ParseInt(digits[numOfDigits-1], 10, 8)
if err != nil {
return false
}
checkSum := int64(0)
for i := 0; i < numOfDigits-1; i++ { // we dont need the last one
convertedDigit := ""
if (i+1)%2 == 0 {
d, err := strconv.ParseInt(digits[i], 10, 8)
if err != nil {
return false
}
convertedDigit = strconv.FormatInt(2*d, 10)
} else {
convertedDigit = digits[i]
}
convertedDigits := strings.Split(convertedDigit, "")
for _, c := range convertedDigits {
d, err := strconv.ParseInt(c, 10, 8)
if err != nil {
return false
}
checkSum = checkSum + d
}
}
if (checkSum+checkingDigit)%10 != 0 {
return false
}
return true
}
回答by Milan Mahata
I think this logic is not right because this working only for the specific IMEI no - 490154203237518 not for other IMEI no ...I implement the code also...
我认为这种逻辑是不对的，因为这仅适用于特定的 IMEI 号 - 490154203237518 不适用于其他 IMEI 号……我也实现了代码……
var number = 490154203237518;
var array1 = new Array();
var array2 = new Array();
var specialno = 0 ;
var sum = 0 ;
var finalsum = 0;
var cast = number.toString(10).split('');
var finalnumber = '';
if(cast.length == 15){
for(var i=0,n = cast.length; i<n; i++){
if(i !== 14){
if(i == 0 || i%2 == 0 ){
array1[i] = cast[i];
}else{
array1[i] = cast[i]*2;
}
}else{
specialno = cast[14];
}
}
for(var j=0,m = array1.length; j<m; j++){
finalnumber = finalnumber.concat(array1[j]);
}
while(finalnumber){
finalsum += finalnumber % 10;
finalnumber = Math.floor(finalnumber / 10);
}
contno = (finalsum/10);
finalcontno = Math.round(contno)+1;
check_specialno = (finalcontno*10) - finalsum;
if(check_specialno == specialno){
alert('Imei')
}else{
alert('Not IMEI');
}
}else{
alert('NOT imei - length not matching');
}
//alert(sum);
回答by roelofs
I don't believe there are any built-in ways to authenticate an IMEI number. You would need to verify against a third party database (googling suggests there are a number of such services, but presumably they also get their information from more centralised sources).
我不相信有任何内置方法可以验证 IMEI 号码。您需要对照第三方数据库进行验证（谷歌搜索表明有许多此类服务，但据推测它们也从更集中的来源获取信息）。
