是否可以縮短此代碼？

Question

我並不是Java的初學者，但我也不是專家。 有什么方法可以縮短此代碼，使其占用更少的空間並可能更少的行數？

JOptionPane.showMessageDialog(null, "The student's names are: " 
  + roster[0][0] + " " + roster[1][0] + ", " 
  + roster[0][1] + " " + roster[1][1] + ", " 
  + roster[0][2] + " " + roster[1][2] + ", and " 
  + roster[0][3] + " " + roster[1][3] + ".");

Answer 1

讓我們逐步重構代碼

1. 引入變量以消除重復。

    T[] col1 = roster[0]; T[] col2 = roster[1]; String content = col1[0] + " " + col2[0] + ", " + col1[1] + " " + col2[1] + ", " + col1[2] + " " + col2[2] + ", and " + col1[3] + " " + col2[3] + "."; JOptionPane.showMessageDialog(null,"The student's names are: " + content); 

2. 將字符串串聯成多個分配，則除最后一個分配外，所有分配均相同。

    int i = 0; String content = ""; content += col1[i] + " " + col2[i] + ", ";i++; content += col1[i] + " " + col2[i] + ", ";i++; content += col1[i] + " " + col2[i] + ", ";i++; content += "and " + col1[i] + " " + col2[i] + ".";i++; 

3. 使用三元運算符使多重分配保持一致。

    int i = 0; String content = ""; content+= (i==3?"and ":"") + col1[i]+" "+col2[i] + (i==3?".":", "); i++; content+= (i==3?"and ":"") + col1[i]+" "+col2[i] + (i==3?".":", "); i++; content+= (i==3?"and ":"") + col1[i]+" "+col2[i] + (i==3?".":", "); i++; content+= (i==3?"and ":"") + col1[i]+" "+col2[i] + (i==3?".":", "); i++; 

4. 使用while循環刪除重復項。

    int i = 0; String content = ""; while(i<=3) { content+= (i==3?"and ":"") + col1[i]+" "+col2[i] + (i==3?".":", "); i++; } 

5. 用for循環替換while循環。

    String content = ""; for (int i = 0; i <= 3; i++) { content+= (i==3?"and ":"") + col1[i]+" "+col2[i] + (i==3?".":", "); } 

6. 引入變量以使代碼更具可讀性。

    String content = ""; for (int i = 0; i <= 3; i++) { String prefix = i == 3 ? "and " : ""; String current = col1[i] + " " + col2[i]; String suffix = i == 3 ? "." : ", "; content += prefix + current + suffix; } 

7. 內聯僅使用一次的變量col1和col2 ：

    String content = ""; for (int i = 0; i <= 3; i++) { String prefix = i == 3 ? "and " : ""; String current = roster[0][i] + " " + roster[1][i]; String suffix = i == 3 ? "." : ", "; content += prefix + current + suffix; } 

8. 用常數替換魔術數字3 ，最終代碼如下：

    final int last = 3; String content = ""; for (int i = 0; i <= last; i++) { String prefix = i == last ? "and " : ""; String suffix = i == last ? "." : ", "; String current = roster[0][i] + " " + roster[1][i]; content += prefix + current + suffix; } JOptionPane.showMessageDialog(null, "The student's names are: " + content); 

Answer 2

StringBuilder message = new StringBuilder("The student's names are: ");
for (int i = 0; i < roster[0].length; i++) {
    message
        .append(roster[0][i])
        .append(" ")
        .append(roster[1][i]);
    if (i < roster[0].length - 1)
        message.append(", ");
    if (i == roster[0].length - 2)
        message.append("and ")
}
message.append(".");
JOptionPane.showMessageDialog(null, message.toString());

可能是這樣的。 如您所見，您實際上並沒有保存任何行，但是顯然，由於可以考慮可變長度的花名冊，因此代碼更加靈活。

Answer 3

您可以執行以下操作：最初分配您不想重復的普通語句。 然后遍歷花名冊。

String rosterString= "The student's names are: ";
for(int i=0;i<= roster.length;i++){
    for(int j=0;j<= roster[i].length;j++){
         rosterString += (roster[i][j] + " ");
         if (i == 1 && j < 2) {
             rosterString += ", ";
         }
         else if (i == 1 && j == 2) {
             rosterString += ", and";
         }
         else if (i == 1 && j == 3) {
             rosterString += ".";
         }
         else {
             rosterString += " ";
         }
    }
}

然后將rosterString傳遞給您的方法。

Answer 4

String rosterString = "";
for(int i = 0; i < roster[0].length; i++) {
    rosterString += roster[0][i] + " " + roster[1][i] + ", ";
}

該代碼將創建一個包含名稱和逗號的字符串。 然后，您可以添加在if語句來檢查它是否接近尾聲改變,一個and或.