如何在 Java 中組合兩個對象數組

Question

我有這個從 excel 文件中獲取信息的小腳本。 在收集了我需要的信息后，我想將這兩個數組合並為一個。 那可能嗎？

public Object[][] createData1() throws Exception {
    Object[][] retObjArr1 = data.getTableArray("C:\\Users\\OAH\\Workspaces\\Chrome2\\Testdata2.xls", "Sheet1", "normalCustomer");
    Object[][] retObjArr2 = data.getTableArray("C:\\Users\\OAH\\Workspaces\\Chrome2\\Testdata2.xls", "Sheet2", "langLogin");
    return(retObjArrCombined); //I want to return one array with both arrays
}

Answer 1

您可以使用System.arraycopy方法（是的，全部小寫）。 文檔。 您可以使用java.util.Arrays類對數組做更多的事情。

Answer 2

這個怎么樣？

鏈接到另一個堆棧問題！！！！！！

    public static int[][] append(int[][] a, int[][] b) {
    int[][] result = new int[a.length + b.length][];
    System.arraycopy(a, 0, result, 0, a.length);
    System.arraycopy(b, 0, result, a.length, b.length);
    return result;
}

Answer 3

這是簡單的解決方案：

private static Object[] concatenate(Object[] a, Object[] b) {
    Collection<Object> result = new ArrayList<Object>(a.length + b.length);
    for (Object val : a) {
        result.add(val);
    }
    for (Object val : b) {
        result.add(val);            
    }
    return result.toArray();
} 

Answer 4

其他方式：

Object[][] one = {{1, 2, 3}, {4, 5}, {6}};
Object[][] two = {{7, 8}, {9}};
List<Object[]> holder = new ArrayList<>();

Collections.addAll(holder, one);
Collections.addAll(holder, two);

Object[][] result = holder.toArray(new Object[holder.size()][]);

Answer 5

以下代碼完成您的任務：

Object[][] retObjArr1 = { { "a00", "a01" }, { "a10", "a11" },
        { "a20", "a21" } };
Object[][] retObjArr2 = { { "b00", "b01" }, { "b10", "b11" },
        { "b20", "b21" } };

List<Object[][]> list = new ArrayList<Object[][]>();
list.add(retObjArr1);
list.add(retObjArr2);

int totalRow = 0;
for (int all = 0; all < list.size(); all++) {
    totalRow += list.get(all).length;
}
Object[][] retObjArrCombined = new Object[totalRow][];
int rowCount = 0;
for (int all = 0; all < list.size(); all++) {
    Object[][] objects = list.get(all);
    for (int i = 0; i < objects.length; i++) {
        retObjArrCombined[rowCount] = objects[i];
        rowCount++;
    }
}
for (int i = 0; i < retObjArrCombined.length; i++) {
    for (int j = 0; j < retObjArrCombined[i].length; j++) {
        System.out.println("value at :(" + i + "," + j + ") is:"
                + retObjArrCombined[i][j]);
    }
}

在此代碼中， Object[][] retObjArrCombined包含從retObjArr1 、 retObjArr2等復制的所有數組......並打印以下輸出：

value at :(0,0) is:a00
value at :(0,1) is:a01
value at :(1,0) is:a10
value at :(1,1) is:a11
value at :(2,0) is:a20
value at :(2,1) is:a21
value at :(3,0) is:b00
value at :(3,1) is:b01
value at :(4,0) is:b10
value at :(4,1) is:b11
value at :(5,0) is:b20
value at :(5,1) is:b21

Answer 6

單行流，無需計算上拉
是的，它會慢幾毫秒

Object[][] combi = Stream.concat( Arrays.stream( retObjArr1 ), Arrays.stream( retObjArr2 ) ).toArray( Object[][]::new );