这是我的第一篇文章,我只是Java的新手,很抱歉,如果它不是最新的。
我一直在用Java编写一个基于文本的冒险游戏,而我的代码在一个地方让我失望 - 解析器。没有错误,它只是不起作用。它接受输入但不做任何事情。它非常简单,看起来像这样:
public static void getInput(){
System.out.print(">>"); //print cue for input
String i = scan.nextLine(); //get (i)nput
String[] w = i.split(" "); //split input into (w)ords
List words = Arrays.asList(w); //change to list format
test(words);
}
测试方法只是使用if(words.contains("<word>"))
在列表中搜索某些单词。
代码有什么问题,如何改进?
答案 0 :(得分:1)
如何保持数组并使用类似的东西:
String[] word_list = {"This","is","an","Array"}; //An Array in your fault its 'w'
for (int i = 0;i < word_list.length;i++) { //Running trough all Elements
System.out.println(word_list[i]);
if (word_list[i].equalsIgnoreCase("This")) {
System.out.println("This found!");
}
if (word_list[i].equalsIgnoreCase("is")) {
System.out.println("is found!");
}
if (word_list[i].equalsIgnoreCase("an")) {
System.out.println("an found!");
}
if (word_list[i].equalsIgnoreCase("Array")) {
System.out.println("Array found!");
}
if (word_list[i].equalsIgnoreCase("NotExistant")) { //Wont be found
System.out.println("NotExistant found!");
}
}
您将获得以下输出:
This found!
is found!
an found!
Array found!
如您所见,您根本无需将其转换为List!
答案 1 :(得分:0)
我将如何做到这一点:
public class ReadInput {
private static void processInput (List <String> words)
{
if (words.contains ("foo"))
System.out.println ("Foo!");
else if (words.contains ("bar"))
System.out.println ("Bar!");
}
public static void readInput () throws Exception
{
BufferedReader reader =
new BufferedReader (
new InputStreamReader (System.in));
String line;
while ((line = reader.readLine ()) != null)
{
String [] words = line.split (" ");
processInput (Arrays.asList (words));
}
}
public static void main (String [] args) throws Exception
{
readInput ();
}
}
示例会话:
[in] hello world
[in] foo bar
[out] Foo!
[in] foobar bar foobar
[out] Bar!