Question

我在过去一周遇到麻烦，试图解决如何检查Id是否已经添加到阵列中，如果有，请阻止用户输入其余数据，我将非常感谢一些帮助有这个。

我的代码：

class CustomerUse {
   public static int checkId(Customer theArray[], int noOfValues,
         String searchId) {
      int idCheck = 0;
      int step = 0;
      while (step < noOfValues) {
         step++;
         if ((step < noOfValues) && (theArray[step].getId().equals(searchId))) {
            idCheck = -1;
         }
      }
      return idCheck;
   }

   public static void listAllNames(Customer[] names, int NoOfElements) {
      int index;
      String result;
      for (index = 0; index < NoOfElements; index++) {
         System.out.println(names[index]);
      }
   }

   public static int findPlace(Customer theArray[], int balance, int noOfValues) {
      int step;
      int place;
      step = 0;
      while ((step < noOfValues) && (balance < theArray[step].getBalance())) {
         step++;
      }
      place = step;
      return place;
   }

   public static int addOne(Customer theArray[], int place, String Id,
         Customer theObject, int noOfValues) {
      int step;
      if (noOfValues == 0) {
         theArray[0] = theObject;
         noOfValues++;
      } else {
         for (step = noOfValues - 1; step >= place; step--) {
            theArray[step + 1] = theArray[step];
         }
         theArray[place] = theObject;
         noOfValues++;
      }
      return noOfValues;
   }

   public static int deleteName(Customer theArray[], int noOfElements) {
      String searchId;
      int step;
      int whichOne = 0;
      System.out.println("Enter Id of customer to be deleted ");
      searchId = EasyIn.getString();
      step = 0;
      while ((step < noOfElements)
            && !(theArray[step].getId().equals(searchId))) {
         step++;
      }
      if (step < noOfElements) {
         whichOne = step;
         for (step = whichOne; step < noOfElements - 1; step++) {
            theArray[step] = theArray[step + 1];
         }
         noOfElements--;
      } else {
         System.out.println(" Sorry this customer doesn't exist");
      }
      return noOfElements;
   }

   public static void main(String[] args) {
      Customer empArray[];
      int index;
      int noOfElements;
      String searchName;
      Customer tempObject;
      int step;
      int option = 0;
      int place;
      String newName = "";
      String newId;
      int newBalance;
      int checkID = 0;
      empArray = new Customer[100000];
      noOfElements = 0;
      System.out
            .println("\n 1. Add Customer \n 2. Delete Customer \n 3. List all Customers \n 4. Exit");
      System.out.print("Enter Option ");
      option = EasyIn.getInt();
      while (option != 4) {
         if (option == 1) {
            System.out.print("Enter ID ");
            newId = EasyIn.getString();
            System.out.print("Enter name ");
            newName = EasyIn.getString();
            System.out.print("Enter Balance ");
            newBalance = EasyIn.getInt();
            System.out.println();
            tempObject = new Customer(newName, newId, newBalance);
            place = findPlace(empArray, newBalance, noOfElements);
            noOfElements = addOne(empArray, place, newId, tempObject,
                  noOfElements);
            checkID = checkId(empArray, noOfElements, newId);
            if (checkID == -1) {
               System.out.println("Sorry this ID already exists");
               checkID = 0;
            }
         } else if (option == 2) {
            noOfElements = deleteName(empArray, noOfElements);
         } else if (option == 3) {
            listAllNames(empArray, noOfElements);
         }
         System.out.print("Enter Option ");
         option = EasyIn.getInt();
      }
   }
}

Answer 1

这有效地跳过索引0处的数组条目：

while (step < noOfValues)
{

    step++;

更好：

public static int checkId(Customer theArray[], int noOfValues, String searchId){
    for (int step = 0; step < noOfValues; ++step ){
        if( theArray[step].getId().equals(searchId) )
        {
            return step;
        }   
    }
    return -1; 
}

未找到时返回-1，否则返回索引。

我没有检查你的代码的其余部分，只是看到应该放弃使用数组而不是List。

ID检查对象数组

1 个答案: