blocks|key|5938078|text|没有给出完整的答案，因为这看起来像一个家庭作业//学习问题？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5938079|从可取到不可取：|5938080|5938081|您可以有一个数组或单词列表，并返回数组或列表中的第“display”项？|unordered-list-item|5938082|您还可以使用switch/case方法，对与显示编号对应的单词进行硬编码。|5938083|您还可以使用较大的if/elseif/elsif格式。|5938084|5938085|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|R|8|@]|9|@]|A|$]]|$1|H|3|I|5|G|7|S|8|@]|9|@]|A|$]]|$1|J|3|K|5|G|7|T|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

Not giving complete answers, as this looks like a homework // learning question?

From desirable to undesirable:

<ul>
<li>You could have an array or list of words, and return the "display"th
item in the array or list?</li>
<li>You could also use a switch/case method and hardcode the words that
correspond with the display number.</li>
<li>You could also use a big if/elseif/elsif format.</li>
</ul>

blocks|key|2409290|text|试一试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2409291|public+class+Main{
++++private+static+String[]+words+=+new+String[]+{"one",+"two",+"three",+"four",+"five",+"six",+"seven",+"eight",+"nine",+"ten"};
++++public+static+void+tenWords(int+display){
++++++++++++System.out.println(words[display]);
++++}

++++public+static+void+main(String[]+args)+{

++++++++for(int+i=0;i<10;i%2B%2B){
++++++++++++tenWords(i);
++++++++}
++++}
}|code-block|syntax|javascript|2409292|冰|2409293|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

just try that:

<pre><code>public class Main{
 private static String[] words = new String[] {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten"};
 public static void tenWords(int display){
 System.out.println(words[display]);
 }

 public static void main(String[] args) {

 for(int i=0;i&lt;10;i++){
 tenWords(i);
 }
 }
}
</code></pre>

ice

blocks|key|4522219|text|您可以通过使用任何循环来反复调用main方法(您的偏好)，但我使用if语句来调用main方法。以下是我的示例代码:将此代码用作reference..you会发现它很方便：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4522220|import+java.util.Scanner;

public+class+NewTest+{

public+static+void+main(String[]+args)+{
++++Scanner+src+=+new+Scanner(System.in);
++++System.out.println("Enter+the+Value:");
++++int+a+=+src.nextInt();
++++char+result;

++++if+(a+>=+90)+{
++++++++result+=+'A';
++++}+

++++else+if+(a+>=+80)+{
++++++++result+=+'B';

++++}+
++++else+if+(a+>=+70)+{
++++++++result+=+'C';
++++}+
++++else+if+(a+>=+60)+{
++++++++result+=+'D';
++++}

++++else+{
++++++++result+=+'F';
++++}

++++if(result+==+0){

++++++++System.out.println("Do+Nothing");
++++}

++++else{

++++++++NewTest+i+=+new+NewTest();
++++++++if(i!=+null){

++++++++++++System.out.println(result);
++++++++}
++++++++++++++++++++//Here+it+goes+to+the+main+method+again+and+runs+it.
++++++++i.main(args);


++++}

}

}|code-block|syntax|javascript|4522221|希望这对你有用...:)|4522222|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You can call the main method again and again by using any of the loops(your preference), but I used if statement to call the main method. Here's my sample code:
Use this as a reference..you will find it handy:

<pre><code>import java.util.Scanner;

public class NewTest {

public static void main(String[] args) {
 Scanner src = new Scanner(System.in);
 System.out.println("Enter the Value:");
 int a = src.nextInt();
 char result;

 if (a &gt;= 90) {
 result = 'A';
 } 

 else if (a &gt;= 80) {
 result = 'B';

 } 
 else if (a &gt;= 70) {
 result = 'C';
 } 
 else if (a &gt;= 60) {
 result = 'D';
 }

 else {
 result = 'F';
 }

 if(result == 0){

 System.out.println("Do Nothing");
 }

 else{

 NewTest i = new NewTest();
 if(i!= null){

 System.out.println(result);
 }
 //Here it goes to the main method again and runs it.
 i.main(args);


 }

}

}
</code></pre>

Hope this works for you... :)

blocks|key|2409322|text|这个怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2409323|public+class+PepsiMaxIsBetterThanDietCoke{

++public+static+void+main(String[]+args){
++String+[]+Words+=+++++{+"horse",+"Cow",+"bullet",+"jenifer",+"maypole",+"dumbbell",+"dog",+"playstaion"++++++++,+"xbox",+"ciggerette"};



for+(String+x+:+Words)
++System.out.println(x);


++++++++}
++++}|code-block|syntax|javascript|2409324|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

how about this

<pre><code>public class PepsiMaxIsBetterThanDietCoke{

 public static void main(String[] args){
 String [] Words = { "horse", "Cow", "bullet", "jenifer", "maypole", "dumbbell", "dog", "playstaion" , "xbox", "ciggerette"};



for (String x : Words)
 System.out.println(x);


 }
 }
</code></pre>

Alright, So I'm learning method in Java, I have to call on a method 10 times to display ten different words (I already have the for loop to call on the method). I just can't figure out how to get it to have 10 different words. This is what I have so far. I hate asking for help so much, but I've been stumped for over a day now.

<pre><code>public static void tenWords(int display){

}

public static void main(String[] args) {

 for(int i=0;i&lt;10;i++){
 tenWords(i);
 }

}
</code></pre>

Calling a method multiple times

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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好了，我正在学习Java中的方法，我必须调用一个方法10次才能显示10个不同的单词(我已经有了for循环来调用这个方法)。我只是想不出怎么让它有10个不同的单词。这就是我到目前为止所拥有的。我讨厌这么多的求助，但我已经被难住了一天多了。public static void tenWords(int display){}...

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