如何解决我已在WebMatrix中使用RazorC#语法将字符串拆分为单个字母,现在如何用结果填充数组?
| 任务:显示手牌(13张牌)。这只手来自于格式为KQT5.KJ873..AJ52的数据库,其中西服顺序为黑桃,红心,钻石球杆,句点用于分隔西服。我希望创建这只手的2D数组,即 [S,K] [S,Q] [S,T] [S,5] [H,K] [H,J] [H,8] [H,7] [H,3] (钻石中有空隙) [C,A] [C,J] [C,5] [C,2] 到目前为止,我在WebMatrix中使用Razor(C#)的代码是@{ string westHand = \"KQT5.KJ873..AJ52\";
foreach (string subString2 in westHand.Split(\'.\')) {
@subString2 <br />
foreach (char c in subString2){
@c <br />
}
}}
输出是
KQT5
ķ
问
Ť
5
KJ873
ķ
Ĵ
8
7
3
AJ52
一种
Ĵ
5
2
现在将各个卡分开。就像我上面说的,我想把它放到一个二维数组中:
string[,] handData = new string[12,12]
嘿,如果我能算出如何将数字放入一维数组中,我会很高兴。
编辑:如下所述,所需数组的尺寸应为[13,2],即13行乘2列。
解决方法
编辑:
好的,我想这正是您要的。最后,手包含一个字符数组:hand [纸牌索引,0-12] [0是花色,1是纸牌]
char[] suits = { \'S\',\'H\',\'D\',\'C\' };
char[,] hand = new char[13,2];
string westHand = \"KQT5.KJ873..AJ52\";
String output = new String();
int currentSuit = 0; //Iterator for suits (0-4)
int currentCard = 0; //Current # of card from hand (0-12)
foreach (string suitString in westHand.Split(\'.\')) {
foreach (char cardChar in suitString){
hand[currentCard,0] = suits[currentSuit];
hand[currentCard,1] = cardChar;
currentCard++;
}
currentSuit++;
}
for(int x = 0; x < 13; x++)
{
output += \"[\" + hand[x,0] + \",\" + hand[x,1] + \"]\";
}
}
输出值:
[S,K][S,Q][S,T][S,5][H,K][H,J][H,8][H,7][H,3][C,A][C,J][C,5][C,2]
上一个答案,以防万一您仍然需要它:
我认为这是您正在尝试做的事情。这只是直接的C#,但是使用类,因为这是一种面向对象的语言。 :)
char[] suits = { \'S\',\'C\' };
String output = new String();
List<Card> hand = new List<Card>();
string westHand = \"KQT5.KJ873..AJ52\";
int currentSuit = 0;
foreach (string suitString in westHand.Split(\'.\')) {
foreach (char cardChar in suitString){
Card newCard = new Card(suits[currentSuit],cardChar);
hand.Add(newCard);
}
currentSuit++;
}
foreach (Card currentCard in hand)
{
output += currentCard.ToString();
}
这是Card类:
public class Card
{
public char suit,type;
public Card(char suit,char type)
{
this.suit = suit;
this.type = type;
}
public String ToString()
{
return \"[\" + this.suit + \",\" + this.type + \"]\";
}
}
输出:
[S,2]
同样,我认为这是您想要的,但是我不确定。让我知道我是否离基地远一点。
, 我不确定是要显示帖子开始时写的卡片,还是要“放入”数组以执行其他操作?但是,仅以所需格式显示它们,以下代码将起作用:
@{ string westHand = \"KQT5.KJ873..AJ52\";
char type = \'S\'; //start with spades
foreach (string subString2 in westHand.Split(\'.\')) {
foreach (char c in subString2){
<text>[@type,@c]</text>
}
switch (type)
{
case \'S\': type = \'H\'; break;
case \'H\': type = \'D\'; break;
case \'D\': type = \'C\'; break;
}
}}
编辑:如果您真的只想将它们放在具有13行2列的数组中,请使用以下代码。 (变量结果包含具有正确值的数组)
string westHand = \"KQT5.KJ873..AJ52\";
char type = \'S\'; //start with spades
string[,] result = new string[westHand.Length - 3,2];
int counter = 0;
foreach (string subString2 in westHand.Split(\'.\'))
{
foreach (char c in subString2)
{
result[counter,0] = type.ToString();
result[counter,1] = c.ToString();
counter++;
}
switch (type)
{
case \'S\': type = \'H\'; break;
case \'H\': type = \'D\'; break;
case \'D\': type = \'C\'; break;
}
}
, 感谢Preli&chrsmtclf。将您的解决方案放入WebMatrix的Razor(C#)语法中,我现在有:
(1)Preli的解决方案;和
@{string westHand = \"KQT5.KJ873..AJ52\";
var num = westHand.Length - 3;
char type = \'S\'; //start with spades
string[,] result = new string[num,2];
int counter = 0;
foreach (string subString2 in westHand.Split(\'.\'))
{
foreach (char card2 in subString2)
{
@: [@type,@card2]
result[counter,0] = type.ToString();
result[counter,1] = card2.ToString();
counter++;
}
switch (type)
{
case \'S\': type = \'H\'; break;
case \'H\': type = \'D\'; break;
case \'D\': type = \'C\'; break;
}
}
}
<br /> You have @num cards. <br />
@for(var i = 0; i < num; i++)
{
@result[i,0] ; @result[i,1] ;<br />
}
(2)chrsmtclf的解决方案
@{ char[] suits = { \'S\',\'C\' };
char[,2];
string westHand = \"KQT5.KJ873..AJ52\";
int currentSuit = 0; //Iterator for suits (0-4)
int currentCard = 0; //Current # of card from hand (0-12)
foreach (string suitString in westHand.Split(\'.\')) {
foreach (char cardChar in suitString){
hand[currentCard,0] = suits[currentSuit];
hand[currentCard,1] = cardChar;
currentCard++;
}
currentSuit++;
}
}
@for(var i = 0; i < 13; i++)
{
@hand[i,0] ; @hand[i,1] ;<br />
}
上面的两个解决方案都提供了二维数组中包含的以下输出:
SK
平方
ST
S5
香港
HJ
H8
H7
H3
认证机构
CJ
C5
C2
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