轉換一組字符串被解析到一個多維數組彩車

Question

我寫了用戶在JTextField中輸入一串值的程序，然後又JTextField中，然後又等。也就是說，每個JTextField中應包含它自己的一套值，我需要一個ArrayList包含這些值的ArrayLists。

我的問題是，它的輸出包含一系列的ArrayLists的ArrayList，所有的人的價值觀是空的。

我稍後也將被添加第三維它，但應該是比較容易的，如果我能得到這個工作。

下面是方法，我嘗試使用的簡化版本的測試：

import java.util.ArrayList; 
import java.util.Arrays; 

public class Test { 
    public static void main(String args[]) { 

    ArrayList<Float> temp = new ArrayList<>();         //holds each float parsed from an ArrayList of strings 
    ArrayList<ArrayList<Float>> output = new ArrayList<>();      //holds each set of floats 
    ArrayList<String> items;             //the string before its parsed into floats 

    String s = "1,2,3,4,5,6";             //testing values 

    for (int i = 0; i <= 10; i++) {            //10 is a random testing number 
     //In the real version s changes here 
     items = new ArrayList<String>(Arrays.asList(s.split("\\s*,\\s*"))); //split strings by comma accounting for the possibility of a space 

     for (String b : items) {            //parse each number in the form of a string into a float, round it, and add it to temp 
      temp.add((float)((long)Math.round(Float.valueOf(b)*10000))/10000); 
     } //End parsing loop 

     output.add(temp);              //put temp in output 
     temp.clear();               //clear temp 
    } //End primary loop 

    System.out.println(output);             //output: [[], [], [], [], [], [], [], [], [], [], []] 
} 

}

Answer 1

這是因爲你一直在清理臨時數組列表。

在for循環中創建ArrayList<Float> temp = new ArrayList<>();並且不清除它。

for (int i = 0; i <= 10; i++) {            //10 is a random testing number 
    //In the real version s changes here 
    items = new ArrayList<String>(Arrays.asList(s.split("\\s*,\\s*"))); //split strings by comma accounting for the possibility of a space 

    // create this inside so you have a new one for each inner list 
    ArrayList<Float> temp = new ArrayList<>(); 
    for (String b : items) {            //parse each number in the form of a string into a float, round it, and add it to temp 
     temp.add((float)((long)Math.round(Float.valueOf(b)*10000))/10000); 
    } //End parsing loop 

    output.add(temp); 

    // don't clear 
    //temp.clear();               //clear temp 
} //End primary loop