这是可行的，但您需要插入自己的错误检查。

public class StringConcatenate {

    public static void main(String[] args){

        // Create two arrays to concatenate and one array to hold both
        String[] arr1 = new String[]{"s","t","r","i","n","g"};
        String[] arr2 = new String[]{"s","t","r","i","n","g"};
        String[] arrBoth = new String[arr1.length+arr2.length];

        // Copy elements from first array into first part of new array
        for(int i = 0; i < arr1.length; i++){
            arrBoth[i] = arr1[i];
        }

        // Copy elements from second array into last part of new array
        for(int j = arr1.length;j < arrBoth.length;j++){
            arrBoth[j] = arr2[j-arr1.length];
        }

        // Print result
        for(int k = 0; k < arrBoth.length; k++){
            System.out.print(arrBoth[k]);
        }

        // Additional line to make your terminal look better at completion!
        System.out.println();
    }
}

它可能不是最有效的，但除了Java自己的API之外，它不依赖其他任何东西。

在Java 8中使用流：

String[] both = Stream.concat(Arrays.stream(a), Arrays.stream(b))
                      .toArray(String[]::new);

或者像这样，使用flatMap：

String[] both = Stream.of(a, b).flatMap(Stream::of)
                      .toArray(String[]::new);

要对泛型类型执行此操作，必须使用反射：

@SuppressWarnings("unchecked")
T[] both = Stream.concat(Arrays.stream(a), Arrays.stream(b)).toArray(
    size -> (T[]) Array.newInstance(a.getClass().getComponentType(), size));

每个答案都是复制数据并创建新阵列。这并不是绝对必要的，如果您的阵列相当大，这绝对不是您想要做的。Java创建者已经知道数组拷贝是浪费的，这就是为什么他们提供System.arrayCopy（）来在我们必须时在Java之外进行这些拷贝的原因。

与其四处复制数据，不如考虑将其保留在原地，并从中提取数据所在的位置。仅仅因为程序员想组织数据位置而复制数据位置并不总是明智的。

// I have arrayA and arrayB; would like to treat them as concatenated
// but leave my damn bytes where they are!
Object accessElement ( int index ) {
     if ( index < 0 ) throw new ArrayIndexOutOfBoundsException(...);
     // is reading from the head part?
     if ( index < arrayA.length )
          return arrayA[ index ];
     // is reading from the tail part?
     if ( index < ( arrayA.length + arrayB.length ) )
          return arrayB[ index - arrayA.length ];
     throw new ArrayIndexOutOfBoundsException(...); // index too large
}

以下是对我有用的：

String[] data=null;
String[] data2=null;
ArrayList<String> data1 = new ArrayList<String>();
for(int i=0; i<2;i++) {
   data2 = input.readLine().split(",");
   data1.addAll(Arrays.asList(data2));
   data= data1.toArray(new String[data1.size()]);
   }

一个与类型无关的变体（已更新-感谢Volley实例化T）：

@SuppressWarnings("unchecked")
public static <T> T[] join(T[]...arrays) {

    final List<T> output = new ArrayList<T>();

    for(T[] array : arrays) {
        output.addAll(Arrays.asList(array));
    }

    return output.toArray((T[])Array.newInstance(
        arrays[0].getClass().getComponentType(), output.size()));
}

public int[] mergeArrays(int [] a, int [] b) {
    int [] merged = new int[a.length + b.length];
    int i = 0, k = 0, l = a.length;
    int j = a.length > b.length ? a.length : b.length;
    while(i < j) {
        if(k < a.length) {
            merged[k] = a[k];
            k++;
        }
        if((l - a.length) < b.length) {
            merged[l] = b[l - a.length];
            l++;
        }
        i++;
    }
    return merged;
}