Generating unique alphanumeric ID's of a fixed length

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I am needing to generate unique id's for users. The 'unique' part is the problem for me.



The 13 character id's which have a format of the following:



0 + random int with range of 1-9 + random letter + random 3 digit number + string 98XX001 For example: 04X90398XX001 --> (0) (4) (X) (903) (98XX001)



This is the code I currently have for generating the ID's:



package generation;

import java.util.ArrayList;
import java.util.List;
import java.util.Random;

public class id
private static List<String> generatedIDs = new ArrayList<String>();
public static void main(String args)
generateIDs(1200);


public static void generateIDs(int numOfUsers)
for (int i = 0; i < numOfUsers; i++)
Random r = new Random();
int first = r.nextInt((9 - 1) + 1) + 1;
char second = (char) (r.nextInt(26) + 'A');
int third = r.nextInt((899 - 100) + 100) + 100;

String id = "0" + first + second + third + "98XX001";
generatedIDs.add(id);
System.out.println(id);






Like I said, I am pretty clueless on how to make sure that they are unique.







share|improve this question

























    up vote
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    down vote

    favorite












    I am needing to generate unique id's for users. The 'unique' part is the problem for me.



    The 13 character id's which have a format of the following:



    0 + random int with range of 1-9 + random letter + random 3 digit number + string 98XX001 For example: 04X90398XX001 --> (0) (4) (X) (903) (98XX001)



    This is the code I currently have for generating the ID's:



    package generation;

    import java.util.ArrayList;
    import java.util.List;
    import java.util.Random;

    public class id
    private static List<String> generatedIDs = new ArrayList<String>();
    public static void main(String args)
    generateIDs(1200);


    public static void generateIDs(int numOfUsers)
    for (int i = 0; i < numOfUsers; i++)
    Random r = new Random();
    int first = r.nextInt((9 - 1) + 1) + 1;
    char second = (char) (r.nextInt(26) + 'A');
    int third = r.nextInt((899 - 100) + 100) + 100;

    String id = "0" + first + second + third + "98XX001";
    generatedIDs.add(id);
    System.out.println(id);






    Like I said, I am pretty clueless on how to make sure that they are unique.







    share|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I am needing to generate unique id's for users. The 'unique' part is the problem for me.



      The 13 character id's which have a format of the following:



      0 + random int with range of 1-9 + random letter + random 3 digit number + string 98XX001 For example: 04X90398XX001 --> (0) (4) (X) (903) (98XX001)



      This is the code I currently have for generating the ID's:



      package generation;

      import java.util.ArrayList;
      import java.util.List;
      import java.util.Random;

      public class id
      private static List<String> generatedIDs = new ArrayList<String>();
      public static void main(String args)
      generateIDs(1200);


      public static void generateIDs(int numOfUsers)
      for (int i = 0; i < numOfUsers; i++)
      Random r = new Random();
      int first = r.nextInt((9 - 1) + 1) + 1;
      char second = (char) (r.nextInt(26) + 'A');
      int third = r.nextInt((899 - 100) + 100) + 100;

      String id = "0" + first + second + third + "98XX001";
      generatedIDs.add(id);
      System.out.println(id);






      Like I said, I am pretty clueless on how to make sure that they are unique.







      share|improve this question











      I am needing to generate unique id's for users. The 'unique' part is the problem for me.



      The 13 character id's which have a format of the following:



      0 + random int with range of 1-9 + random letter + random 3 digit number + string 98XX001 For example: 04X90398XX001 --> (0) (4) (X) (903) (98XX001)



      This is the code I currently have for generating the ID's:



      package generation;

      import java.util.ArrayList;
      import java.util.List;
      import java.util.Random;

      public class id
      private static List<String> generatedIDs = new ArrayList<String>();
      public static void main(String args)
      generateIDs(1200);


      public static void generateIDs(int numOfUsers)
      for (int i = 0; i < numOfUsers; i++)
      Random r = new Random();
      int first = r.nextInt((9 - 1) + 1) + 1;
      char second = (char) (r.nextInt(26) + 'A');
      int third = r.nextInt((899 - 100) + 100) + 100;

      String id = "0" + first + second + third + "98XX001";
      generatedIDs.add(id);
      System.out.println(id);






      Like I said, I am pretty clueless on how to make sure that they are unique.









      share|improve this question










      share|improve this question




      share|improve this question









      asked Jun 21 at 9:56









      Alex Ritchie

      437




      437




















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          A large part of your id (8 of 13 characters) is actually static.



          That leaves you with 260.000 variations (10 (0-9) x 26 (A-Z) x 1000 (0-999)) of ids to assign to your users. After that, you run out of ids. Will that be enough?



          In your current code there is not much you can do besides checking wether the id you generated is already in the generatedIDs list and if it is, try again.



          Is there a purpose behind generating the ids randomly? Because if they are only used to identify users, it would make more sense to increment them one number at a time and just remember the last id you gave to a user.



          User 1: (0)(0)(A)(000)98XX001
          User 2: (0)(0)(A)(001)98XX001
          User 5000: (0)(0)(E)(999)98XX001
          User 80543: (0)(3)(B)(542)98XX001
          User 260000: (0)(9)(Z)(999)98XX001


          Here would be the code for that. You hand the function the id you would like to start from and how many ids from that start-id it should generate.



          import java.util.ArrayList;
          import java.util.List;

          public class MainClass

          public static void main(String args)

          TestClass gen = new TestClass();
          gen.generateIDs("04X99798XX001", 10);



          public class TestClass

          public static Integer letterToInt(char letter)
          return letter - 'A' + 1;


          public static char intToLetter(int number)
          return (char) (90 - (26 - number));


          private static List<String> generatedIDs = new ArrayList<String>();

          public static void generateIDs(String lastID, Integer numOfUsers)
          String first = lastID.substring(1, 2);
          char second = lastID.substring(2, 3).charAt(0);
          String third = lastID.substring(3, 6);

          Integer firstNum = Integer.parseInt(first);
          Integer secondNum = letterToInt(second);
          Integer thirdNum = Integer.parseInt(third);

          for (Integer i = new Integer(0); i < numOfUsers; i++)
          thirdNum = new Integer(thirdNum.intValue() + 1);

          if (thirdNum > 999)
          thirdNum = new Integer(0);
          secondNum = new Integer(secondNum.intValue() + 1);


          if (secondNum > 26)
          secondNum = new Integer(1);
          firstNum = new Integer(firstNum.intValue() + 1);


          if (firstNum > 9)
          // Ran out of IDs, do something.
          return;


          first = String.valueOf(firstNum);
          second = intToLetter(secondNum);
          third = String.valueOf(thirdNum);

          while (third.length() < 3)
          third = "0" + third;


          String nextID = "0" + first + intToLetter(secondNum) + String.valueOf(third) + "98XX001";
          generatedIDs.add(nextID);

          System.out.println(nextID);





          The result looks like this.



          04X99898XX001
          04X99998XX001
          04Y00098XX001
          04Y00198XX001
          04Y00298XX001
          04Y00398XX001
          04Y00498XX001
          04Y00598XX001
          04Y00698XX001
          04Y00798XX001





          share|improve this answer























          • there are better options for translating letters from/to numbers then switch with 26 cases (like math operation on char)
            – Sharon Ben Asher
            Jun 21 at 13:08











          • @SharonBenAsher I'm sure there are and I think he even did that in his original question. I'm not much of a Java guy, but feel free to edit the answer.
            – Endauriel
            Jun 21 at 13:13











          • I will. and if you are not much of a Java guy, I suggest you write the solution as an abstract description. the starting text is clear enough without the need for explicit code
            – Sharon Ben Asher
            Jun 21 at 13:19










          • @SharonBenAsher Sorry, just trying to help. I've read into it and edited the code.
            – Endauriel
            Jun 21 at 13:31










          Your Answer




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          up vote
          0
          down vote













          A large part of your id (8 of 13 characters) is actually static.



          That leaves you with 260.000 variations (10 (0-9) x 26 (A-Z) x 1000 (0-999)) of ids to assign to your users. After that, you run out of ids. Will that be enough?



          In your current code there is not much you can do besides checking wether the id you generated is already in the generatedIDs list and if it is, try again.



          Is there a purpose behind generating the ids randomly? Because if they are only used to identify users, it would make more sense to increment them one number at a time and just remember the last id you gave to a user.



          User 1: (0)(0)(A)(000)98XX001
          User 2: (0)(0)(A)(001)98XX001
          User 5000: (0)(0)(E)(999)98XX001
          User 80543: (0)(3)(B)(542)98XX001
          User 260000: (0)(9)(Z)(999)98XX001


          Here would be the code for that. You hand the function the id you would like to start from and how many ids from that start-id it should generate.



          import java.util.ArrayList;
          import java.util.List;

          public class MainClass

          public static void main(String args)

          TestClass gen = new TestClass();
          gen.generateIDs("04X99798XX001", 10);



          public class TestClass

          public static Integer letterToInt(char letter)
          return letter - 'A' + 1;


          public static char intToLetter(int number)
          return (char) (90 - (26 - number));


          private static List<String> generatedIDs = new ArrayList<String>();

          public static void generateIDs(String lastID, Integer numOfUsers)
          String first = lastID.substring(1, 2);
          char second = lastID.substring(2, 3).charAt(0);
          String third = lastID.substring(3, 6);

          Integer firstNum = Integer.parseInt(first);
          Integer secondNum = letterToInt(second);
          Integer thirdNum = Integer.parseInt(third);

          for (Integer i = new Integer(0); i < numOfUsers; i++)
          thirdNum = new Integer(thirdNum.intValue() + 1);

          if (thirdNum > 999)
          thirdNum = new Integer(0);
          secondNum = new Integer(secondNum.intValue() + 1);


          if (secondNum > 26)
          secondNum = new Integer(1);
          firstNum = new Integer(firstNum.intValue() + 1);


          if (firstNum > 9)
          // Ran out of IDs, do something.
          return;


          first = String.valueOf(firstNum);
          second = intToLetter(secondNum);
          third = String.valueOf(thirdNum);

          while (third.length() < 3)
          third = "0" + third;


          String nextID = "0" + first + intToLetter(secondNum) + String.valueOf(third) + "98XX001";
          generatedIDs.add(nextID);

          System.out.println(nextID);





          The result looks like this.



          04X99898XX001
          04X99998XX001
          04Y00098XX001
          04Y00198XX001
          04Y00298XX001
          04Y00398XX001
          04Y00498XX001
          04Y00598XX001
          04Y00698XX001
          04Y00798XX001





          share|improve this answer























          • there are better options for translating letters from/to numbers then switch with 26 cases (like math operation on char)
            – Sharon Ben Asher
            Jun 21 at 13:08











          • @SharonBenAsher I'm sure there are and I think he even did that in his original question. I'm not much of a Java guy, but feel free to edit the answer.
            – Endauriel
            Jun 21 at 13:13











          • I will. and if you are not much of a Java guy, I suggest you write the solution as an abstract description. the starting text is clear enough without the need for explicit code
            – Sharon Ben Asher
            Jun 21 at 13:19










          • @SharonBenAsher Sorry, just trying to help. I've read into it and edited the code.
            – Endauriel
            Jun 21 at 13:31














          up vote
          0
          down vote













          A large part of your id (8 of 13 characters) is actually static.



          That leaves you with 260.000 variations (10 (0-9) x 26 (A-Z) x 1000 (0-999)) of ids to assign to your users. After that, you run out of ids. Will that be enough?



          In your current code there is not much you can do besides checking wether the id you generated is already in the generatedIDs list and if it is, try again.



          Is there a purpose behind generating the ids randomly? Because if they are only used to identify users, it would make more sense to increment them one number at a time and just remember the last id you gave to a user.



          User 1: (0)(0)(A)(000)98XX001
          User 2: (0)(0)(A)(001)98XX001
          User 5000: (0)(0)(E)(999)98XX001
          User 80543: (0)(3)(B)(542)98XX001
          User 260000: (0)(9)(Z)(999)98XX001


          Here would be the code for that. You hand the function the id you would like to start from and how many ids from that start-id it should generate.



          import java.util.ArrayList;
          import java.util.List;

          public class MainClass

          public static void main(String args)

          TestClass gen = new TestClass();
          gen.generateIDs("04X99798XX001", 10);



          public class TestClass

          public static Integer letterToInt(char letter)
          return letter - 'A' + 1;


          public static char intToLetter(int number)
          return (char) (90 - (26 - number));


          private static List<String> generatedIDs = new ArrayList<String>();

          public static void generateIDs(String lastID, Integer numOfUsers)
          String first = lastID.substring(1, 2);
          char second = lastID.substring(2, 3).charAt(0);
          String third = lastID.substring(3, 6);

          Integer firstNum = Integer.parseInt(first);
          Integer secondNum = letterToInt(second);
          Integer thirdNum = Integer.parseInt(third);

          for (Integer i = new Integer(0); i < numOfUsers; i++)
          thirdNum = new Integer(thirdNum.intValue() + 1);

          if (thirdNum > 999)
          thirdNum = new Integer(0);
          secondNum = new Integer(secondNum.intValue() + 1);


          if (secondNum > 26)
          secondNum = new Integer(1);
          firstNum = new Integer(firstNum.intValue() + 1);


          if (firstNum > 9)
          // Ran out of IDs, do something.
          return;


          first = String.valueOf(firstNum);
          second = intToLetter(secondNum);
          third = String.valueOf(thirdNum);

          while (third.length() < 3)
          third = "0" + third;


          String nextID = "0" + first + intToLetter(secondNum) + String.valueOf(third) + "98XX001";
          generatedIDs.add(nextID);

          System.out.println(nextID);





          The result looks like this.



          04X99898XX001
          04X99998XX001
          04Y00098XX001
          04Y00198XX001
          04Y00298XX001
          04Y00398XX001
          04Y00498XX001
          04Y00598XX001
          04Y00698XX001
          04Y00798XX001





          share|improve this answer























          • there are better options for translating letters from/to numbers then switch with 26 cases (like math operation on char)
            – Sharon Ben Asher
            Jun 21 at 13:08











          • @SharonBenAsher I'm sure there are and I think he even did that in his original question. I'm not much of a Java guy, but feel free to edit the answer.
            – Endauriel
            Jun 21 at 13:13











          • I will. and if you are not much of a Java guy, I suggest you write the solution as an abstract description. the starting text is clear enough without the need for explicit code
            – Sharon Ben Asher
            Jun 21 at 13:19










          • @SharonBenAsher Sorry, just trying to help. I've read into it and edited the code.
            – Endauriel
            Jun 21 at 13:31












          up vote
          0
          down vote










          up vote
          0
          down vote









          A large part of your id (8 of 13 characters) is actually static.



          That leaves you with 260.000 variations (10 (0-9) x 26 (A-Z) x 1000 (0-999)) of ids to assign to your users. After that, you run out of ids. Will that be enough?



          In your current code there is not much you can do besides checking wether the id you generated is already in the generatedIDs list and if it is, try again.



          Is there a purpose behind generating the ids randomly? Because if they are only used to identify users, it would make more sense to increment them one number at a time and just remember the last id you gave to a user.



          User 1: (0)(0)(A)(000)98XX001
          User 2: (0)(0)(A)(001)98XX001
          User 5000: (0)(0)(E)(999)98XX001
          User 80543: (0)(3)(B)(542)98XX001
          User 260000: (0)(9)(Z)(999)98XX001


          Here would be the code for that. You hand the function the id you would like to start from and how many ids from that start-id it should generate.



          import java.util.ArrayList;
          import java.util.List;

          public class MainClass

          public static void main(String args)

          TestClass gen = new TestClass();
          gen.generateIDs("04X99798XX001", 10);



          public class TestClass

          public static Integer letterToInt(char letter)
          return letter - 'A' + 1;


          public static char intToLetter(int number)
          return (char) (90 - (26 - number));


          private static List<String> generatedIDs = new ArrayList<String>();

          public static void generateIDs(String lastID, Integer numOfUsers)
          String first = lastID.substring(1, 2);
          char second = lastID.substring(2, 3).charAt(0);
          String third = lastID.substring(3, 6);

          Integer firstNum = Integer.parseInt(first);
          Integer secondNum = letterToInt(second);
          Integer thirdNum = Integer.parseInt(third);

          for (Integer i = new Integer(0); i < numOfUsers; i++)
          thirdNum = new Integer(thirdNum.intValue() + 1);

          if (thirdNum > 999)
          thirdNum = new Integer(0);
          secondNum = new Integer(secondNum.intValue() + 1);


          if (secondNum > 26)
          secondNum = new Integer(1);
          firstNum = new Integer(firstNum.intValue() + 1);


          if (firstNum > 9)
          // Ran out of IDs, do something.
          return;


          first = String.valueOf(firstNum);
          second = intToLetter(secondNum);
          third = String.valueOf(thirdNum);

          while (third.length() < 3)
          third = "0" + third;


          String nextID = "0" + first + intToLetter(secondNum) + String.valueOf(third) + "98XX001";
          generatedIDs.add(nextID);

          System.out.println(nextID);





          The result looks like this.



          04X99898XX001
          04X99998XX001
          04Y00098XX001
          04Y00198XX001
          04Y00298XX001
          04Y00398XX001
          04Y00498XX001
          04Y00598XX001
          04Y00698XX001
          04Y00798XX001





          share|improve this answer















          A large part of your id (8 of 13 characters) is actually static.



          That leaves you with 260.000 variations (10 (0-9) x 26 (A-Z) x 1000 (0-999)) of ids to assign to your users. After that, you run out of ids. Will that be enough?



          In your current code there is not much you can do besides checking wether the id you generated is already in the generatedIDs list and if it is, try again.



          Is there a purpose behind generating the ids randomly? Because if they are only used to identify users, it would make more sense to increment them one number at a time and just remember the last id you gave to a user.



          User 1: (0)(0)(A)(000)98XX001
          User 2: (0)(0)(A)(001)98XX001
          User 5000: (0)(0)(E)(999)98XX001
          User 80543: (0)(3)(B)(542)98XX001
          User 260000: (0)(9)(Z)(999)98XX001


          Here would be the code for that. You hand the function the id you would like to start from and how many ids from that start-id it should generate.



          import java.util.ArrayList;
          import java.util.List;

          public class MainClass

          public static void main(String args)

          TestClass gen = new TestClass();
          gen.generateIDs("04X99798XX001", 10);



          public class TestClass

          public static Integer letterToInt(char letter)
          return letter - 'A' + 1;


          public static char intToLetter(int number)
          return (char) (90 - (26 - number));


          private static List<String> generatedIDs = new ArrayList<String>();

          public static void generateIDs(String lastID, Integer numOfUsers)
          String first = lastID.substring(1, 2);
          char second = lastID.substring(2, 3).charAt(0);
          String third = lastID.substring(3, 6);

          Integer firstNum = Integer.parseInt(first);
          Integer secondNum = letterToInt(second);
          Integer thirdNum = Integer.parseInt(third);

          for (Integer i = new Integer(0); i < numOfUsers; i++)
          thirdNum = new Integer(thirdNum.intValue() + 1);

          if (thirdNum > 999)
          thirdNum = new Integer(0);
          secondNum = new Integer(secondNum.intValue() + 1);


          if (secondNum > 26)
          secondNum = new Integer(1);
          firstNum = new Integer(firstNum.intValue() + 1);


          if (firstNum > 9)
          // Ran out of IDs, do something.
          return;


          first = String.valueOf(firstNum);
          second = intToLetter(secondNum);
          third = String.valueOf(thirdNum);

          while (third.length() < 3)
          third = "0" + third;


          String nextID = "0" + first + intToLetter(secondNum) + String.valueOf(third) + "98XX001";
          generatedIDs.add(nextID);

          System.out.println(nextID);





          The result looks like this.



          04X99898XX001
          04X99998XX001
          04Y00098XX001
          04Y00198XX001
          04Y00298XX001
          04Y00398XX001
          04Y00498XX001
          04Y00598XX001
          04Y00698XX001
          04Y00798XX001






          share|improve this answer















          share|improve this answer



          share|improve this answer








          edited Jun 21 at 14:17


























          answered Jun 21 at 12:59









          Endauriel

          83




          83











          • there are better options for translating letters from/to numbers then switch with 26 cases (like math operation on char)
            – Sharon Ben Asher
            Jun 21 at 13:08











          • @SharonBenAsher I'm sure there are and I think he even did that in his original question. I'm not much of a Java guy, but feel free to edit the answer.
            – Endauriel
            Jun 21 at 13:13











          • I will. and if you are not much of a Java guy, I suggest you write the solution as an abstract description. the starting text is clear enough without the need for explicit code
            – Sharon Ben Asher
            Jun 21 at 13:19










          • @SharonBenAsher Sorry, just trying to help. I've read into it and edited the code.
            – Endauriel
            Jun 21 at 13:31
















          • there are better options for translating letters from/to numbers then switch with 26 cases (like math operation on char)
            – Sharon Ben Asher
            Jun 21 at 13:08











          • @SharonBenAsher I'm sure there are and I think he even did that in his original question. I'm not much of a Java guy, but feel free to edit the answer.
            – Endauriel
            Jun 21 at 13:13











          • I will. and if you are not much of a Java guy, I suggest you write the solution as an abstract description. the starting text is clear enough without the need for explicit code
            – Sharon Ben Asher
            Jun 21 at 13:19










          • @SharonBenAsher Sorry, just trying to help. I've read into it and edited the code.
            – Endauriel
            Jun 21 at 13:31















          there are better options for translating letters from/to numbers then switch with 26 cases (like math operation on char)
          – Sharon Ben Asher
          Jun 21 at 13:08





          there are better options for translating letters from/to numbers then switch with 26 cases (like math operation on char)
          – Sharon Ben Asher
          Jun 21 at 13:08













          @SharonBenAsher I'm sure there are and I think he even did that in his original question. I'm not much of a Java guy, but feel free to edit the answer.
          – Endauriel
          Jun 21 at 13:13





          @SharonBenAsher I'm sure there are and I think he even did that in his original question. I'm not much of a Java guy, but feel free to edit the answer.
          – Endauriel
          Jun 21 at 13:13













          I will. and if you are not much of a Java guy, I suggest you write the solution as an abstract description. the starting text is clear enough without the need for explicit code
          – Sharon Ben Asher
          Jun 21 at 13:19




          I will. and if you are not much of a Java guy, I suggest you write the solution as an abstract description. the starting text is clear enough without the need for explicit code
          – Sharon Ben Asher
          Jun 21 at 13:19












          @SharonBenAsher Sorry, just trying to help. I've read into it and edited the code.
          – Endauriel
          Jun 21 at 13:31




          @SharonBenAsher Sorry, just trying to help. I've read into it and edited the code.
          – Endauriel
          Jun 21 at 13:31












           

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