Validating number in java

lv_C := lv_C lv_A; lv_D := 10 - (lv_C mod 10); IF (lv_D = 10) THEN lv_D := 0; END IF;-- compare value in lv_D with lv_last_number lv_last_number := CAST(SUBSTR(p_id_number, 13, 1) AS NUMBER); IF (lv_last_number = lv_D) THEN lv_is_valid := 1; lv_rply := TRUE; END IF; RETURN lv_rply; -- ************************************ -- ** Handle an unexpected exception ** -- ************************************ EXCEPTION WHEN lv_length_err THEN lv_rply := FALSE; return lv_rply; WHEN no_data_found THEN lv_rply := FALSE; return lv_rply; --custom exception (place your own methods here) --WHEN OTHERS THEN -- raise_exception(SQLCODE, SQLERRM, 'FN_VALID_SA_ID_NUMBER'); END; Hi there I am a technical consultant and would like to know what the SSS as well as the A (8 & 9 in the id) and the z stands for.Please also confirm whether the the intelligence built in the South African Id number remains the same!!!!This format stores numbers in 64 bits, where the number (the fraction) is stored in bits 0 to 51, the exponent in bits 52 to 62, and the sign in bit 63: By default, Java Script displays numbers as base 10 decimals.

validating number in java-38

NET webservice to validate South African Person ID numbers.

It provides validation of a person ID by calculating the check-digit (digit-13), and extracts the date-of-birth, gender, citizenship, sequence and other information.

I ran a small hot script that would post ID number to the URL and checked relevant returned html text for match.

Recently (probably because elections are passed) the site stopped supplying names, but only confimed whether registered or not... Did you know: The current SA ID Number algorithm was created by my uncle Pieter den Boer (ex Q-Data, Brainware...) while working at IBM in the 1980's.

-- If the result is 2 digits, the last digit is used to compare against the last number in the ID Number.

-- If the answer differs, the ID number is invalid.Regards Jan['validate-id', 'You have entered an invalid identity number', function (v) ] Hi, please can someone help with out with this problem: The second last digit in my driver's licence is DIFFERENT to that in my green ID book. Here's my code: CREATE OR REPLACE FUNCTION CS_VALID_SA_ID_NUMBER ( p_id_number IN VARCHAR2 ) RETURN BOOLEANAS lv_A NUMBER := 0; -- ~Holds Negative Index numbers lv_B NUMBER := 0; -- ~Holds Positive Index numbers lv_C NUMBER; lv_D NUMBER := -1; -- ~The check digit lv_I NUMBER; -- ~Counter Variable lv_is_valid NUMBER; lv_the_number NUMBER; lv_rply BOOLEAN := FALSE; BEGIN--Getting Negative Numbers lv_I := 0 ; IF (lv_I 0) THEN lv_C := lv_C lv_B mod 10; lv_B := lv_B / 10; END IF;lv_C := lv_C lv_A;lv_D := 10 - (lv_C mod 10); IF (lv_D = 10) THEN lv_D := 0; END IF;lv_the_number := CAST(SUBSTR(p_id_number, 13, 1) AS NUMBER); IF lv_the_number = lv_D THEN lv_is_valid := 1; lv_rply := TRUE; END IF; RETURN lv_rply; -- ************************************ -- ** Handle an unexpected exception ** -- ************************************ EXCEPTION /* When this function throws an exception outside its exception block via the exception procedure will it catch its own exception.For example, the last three digits in my driver's licence is 050 --- but in my ID it is 084This causes endless confusion. Id # to the picture and name of the person on it so I know if the person I am dealing with is legitimate or just using an valid ID # with his picture on it . I'm trying to convert Craig Peacocks SQL version to PL/SQL function and not getting anywhere. To avaoid this the following statement is required.Substring(2 * i 1, 1)) Next b *= 2 Dim c As Integer = 0 Do c = b Mod 10 b = Int(b / 10) Loop Until b On Saturday, 28 July 2007, my wife (PG Nosi) and bought a television.We found out that she shares her identity number (7602210298085) with a certain Mr Paul Nkosi.This is a worrying situation, especially with the recent corruption reports that we get regarding the Home Affairs Department.

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