日期:2014-05-18 浏览次数:21127 次
var a = ["红色", "蓝色", "白色"];
var b = ["大", "中", "小 "]
var c = [];
for(var i in a)
for(var j in b)
c.push([a[i], b[j]]);
alert(c);
------解决方案--------------------
string[] arr1 = new string[] { "红色", "蓝色", "白色" };
string[] arr2 = new string[] { "大", "中", "小","超大" };
string[][] arr3 = new string[arr1.Length*arr2.Length][];
for (int i = 0; i < arr1.Length; i++)
{
for (int j = 0; j < arr2.Length; j++)
{
string[] arr = new string[] { arr1[i], arr2[j] };
arr3[i + arr1.Length*j] = arr;
}
}
------解决方案--------------------
下面代码,加上必要using,直接运行,就是你要结果。
class Program
{
static void Main(string[] args)
{
testArray();
}
private static void testArray()
{
object[] a = {"红色", "蓝色", "白色","黑色"};
object[] b = {"大", "中", "小"};
int n = a.Length*b.Length;
object[][] c = new object[n][];
for (int i = 0; i < n; i++)
{
c[i] = new object[2];
c[i][0] = a[i / b.Length];
c[i][1] = b[i % b.Length];
}
System.Console.WriteLine("a = " + Array2String(a));
System.Console.WriteLine("b = " + Array2String(b));
System.Console.WriteLine("c = " + Array2String(c));
System.Console.ReadLine();
}
private static string Array2String(object[] arr){
StringBuilder sb = new StringBuilder();
sb.Append("{");
for (int i = 0;i<arr.Length;i++)
{
object o = arr[i];
if (o is Array)
sb.Append(Array2String((object[])o));
else
sb.Append(o);
if (i < arr.Length - 1)
{
sb.Append(",");
}
}
sb.Append("}");
return sb.ToString();
}
}
------解决方案--------------------
class testArray
{
private string[] str1 = { "红", "黄", "蓝" };
private string[] str2 = { "大", "中", "小" };
private string[,] str3 = null;
public void newArray()
{
int m = str1.Length;
int n = str2.Length;
str3 = new string[m, n];
for (int i = 0; i < str1.Length; i++)
{
for (int j = 0; j < str2.Length; j++)
{
str3[i,j] = str1[i] + str2[j];
}
}
}
public void printNewArray()
{
for (int i = 0; i < str1.Length; i++)
{
for (int j = 0; j < str2.Length; j++)
{
System.Console.WriteLine(str3[i,j]);
}
}
}
static void Main(string[] args)
{
testArray ta = new testArray();
ta.newArray();
ta.printNewArray();
}
------解决方案--------------------