循環，以意外的結果進行迭代

Question

我必須進行循環，但是我嘗試了很多時間才能獲得預期的結果，但是我不知道我在做什么錯。

public class JavaApplication34 {
   int totalResults = 248; //
   int itemsperPage = 10;
   int i = 0;
   int j = 0;
   int count = 0;

   public void makeLoop() {
      for (i = 0; i < 10; i++) {
         System.out.println("nextPage " + i);
         this.count = i + 1;
         for (j = 0; j < 10; j++) {
            System.out.println("Filenummer " + (j + 1));
            this.count = j + 1;
            System.out.println("count " + count);
         }
      }
   }

   public static void main(String[] args) {
      JavaApplication34 myTest = new JavaApplication34();
      myTest.makeLoop();
   }        
}

我應該在控制台上顯示以下結果

• 頁面1，filenummer 1，filenummer 2，filenummer 3 ...，filenummer 10
• 頁2，filenummer 11，filenummer 12，filenummer 13 ...，filenummer20
• 第3頁，依此類推

Answer 1

根據預期的輸出，您也可以更改代碼

for (i = 1; i <= (totalResults/itemsperPage) + 1; i++) {
    System.out.println("nextPage " + i);
    for (; j < i* itemsperPage; j++) {
        if(j > totalResults) {
           break;
        }
        System.out.println("Filenummer " + (j + 1));
    }
}

Answer 2

我現在有以下代碼，aeshang寫道：

公共類JavaApplication34 {

int totalResults = 34;
int itemsperPage = 10;
int i = 0;
int j = 0;

public void makeLoop() {

    for (i = 1; i <= (totalResults / itemsperPage); i++) {
        System.out.println("nextPage " + i);
        for (; j < i * itemsperPage; j++) {
            if (j > totalResults) {
                break;
            }
            System.out.println("Filenummer " + (j + 1));
        }
    }

}

它給我以下輸出

• nexpage 1，文件號1，filnumber 2，....，filnumber 10
• nexpage 2，文件號11，文件號12 ....，filnumber 20
• ...........
• nexpage 3，文件號21，文件號22，... filnumber 30（但應保留到34）

Answer 3

        int totalResults = 34;
        int itemsperPage = 10;
        int i = 0;
        int j = 0;
        int loopSize = (totalResults / itemsperPage);

        public void makeLoop() {
            loopSize = (totalResults % itemsperPage > 0) ? ++loopSize : loopSize;

            for (i = 1; i <= loopSize; i++) {
                System.out.print("nextPage " + i + "...");
                for (; j < i * itemsperPage; j++) {
                    if (j > totalResults) {
                        break;
                    }
                    System.out.print(" Filenummer " + (j + 1));
                }
                System.out.println();

            }

        }

輸出= nextPage 1 ... Filenummer 1 Filenummer 2 Filenummer 3 Filenummer 4 Filenummer 5 Filenummer 6 Filenummer 7 Filenummer 8 Filenummer 9 Filenummer 10 nextPage 2 ... Filenummer 11 Filenummer 12 Filenummer 13 Filenummer 14 Filenummer 15 Filenummer 16 Filenummer 17 Filenummer 18 Filenummer 19 Filenummer 20 nextPage 3 ... Filenummer 21 Filenummer 22 Filenummer 23 Filenummer 24 Filenummer 25 Filenummer 26 Filenummer 27 Filenummer 28 Filenummer 29 Filenummer 30 nextPage 4 ... Filenummer 31 Filenummer 32 Filenummer 33 Filenummer 34 Filenummer 35

Answer 4

int totalResults = 34;
int itemsperPage = 10;
int i = 0;
int j = 1;
int loopSize = (totalResults / itemsperPage);

public void makeLoop() {
    loopSize = (totalResults % itemsperPage > 0) ? ++loopSize : loopSize;

    for (i = 1; i <= loopSize; i++) {
        System.out.print("nextPage " + i + "...");
        for (; j < i * itemsperPage; j++) {
            if (j > totalResults) {
                break;
            }
            System.out.print(" Filenummer " + (j));
        }
        System.out.println();

    }

}

輸出= nextPage 1 ... Filenummer 1 Filenummer 2 Filenummer 3 Filenummer 4 Filenummer 5 Filenummer 6 Filenummer 7 Filenummer 8 Filenummer 9 nextPage 2 ... Filenummer 10 Filenummer 11 Filenummer 12 Filenummer 13 Filenummer 14 Filenummer 15 Filenummer 16 Filenummer 17 Filenummer 18 Filenummer 19 nextPage 3 ... Filenummer 20 Filenummer 21 Filenummer 22 Filenummer 23 Filenummer 24 Filenummer 25 Filenummer 26 Filenummer 27 Filenummer 28 Filenummer 29 nextPage 4 ... Filenummer 30 Filenummer 31 Filenummer 32 Filenummer 33 Filenummer 34