在從數組索引中爲隨機輸出放置字符串時出現問題

Question

遇到從數組索引爲隨機語句輸出輸出字符串的問題。哈希映射很好地捕獲了數組中每個字符串的數量，但是一旦它具有所有信息都會給出這些項目的int顯示。目標是讓用戶在選擇主題 - 動詞 - 形容詞 - 賓語 - 副詞之後選擇表單項，然後進行隨機輸出。 也出於某種原因，當它在命名前遇到了用戶輸入信息時，它會碰到副詞。

/* 
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* and open the template in the editor. 
*/ 
package shoutbox; 
import java.util.HashMap; 
import java.util.Random; 
import java.util.Scanner; 
/** 
* 
* @author alden 
*/ 
public class Shoutbox { 

    /** 
    * 
    */ 
    public int t; 
    public int o; 
    public int r; 
    public int q; 
    public int p; 
public String subjectArray[]=new String[t]; // Intilized array by the size of int y 
public String verbArray[]=new String[o]; 
public String adjectiveArray[]=new String[r]; 
public String objectArray[]=new String[q]; 
public String adverbArray[]=new String[p]; 

public Shoutbox(){ 


} 


    public static void main(String[] args) { 
     // TODO code application logic here 
     Boolean repeat = true; 
     System.out.println(" Please select a option"); 

     HashMap <Integer, String> map = new HashMap<>(); // starts the hashmap 


     map.put(0,"Are you new to programming"); 
     map.put(1, "I think java is a good start"); 
     map.put(2, "It is easy to learn"); 
     map.put(3, "You tube is a great resource"); 
     map.put(4,"Most people start with hello world"); 
     map.put(5,"After this you learn Scanners and getter and setters"); 
     map.put(6,"Next is encapsulation"); 
     map.put(7,"After this is loops"); 
     map.put(8,"Once you get loops work on arrays"); 
     map.put(9,"Next you should look at objects"); 


     map.entrySet().forEach((entry) -> { 
      // for loop for the display of has map 
      int key = entry.getKey(); 
      String value = entry.getValue(); 
      System.out.println(key + " : " + value); 
     }); 
char y; 
while (repeat = true){ // gives options for y or n 
System.out.println("Please enter a value"); 
Scanner in = new Scanner(System.in); 
int n = in.nextInt(); 
String s = map.get(n); 
System.out.println(s); 

System.out.println("Would you like to enter another value. y/n"); 
y = in.next().charAt(0); 

      switch (y) { 
       case 'y': 
        repeat = true; 
        break; 
       case 'Y': 
        repeat = true; 
        break; 
       default: 
        repeat = false; 
        break; 
      } 
break; 

} 
String input1; 

boolean loop = false; 
Scanner in = new Scanner(System.in); 

System.out.printf("How many subjects are there?"); 
int t = in.nextInt(); 
String subjectArray[]=new String[t]; 

System.out.printf("How many verbs are there?"); 
int o = in.nextInt(); 
String verbArray[]=new String[o]; 

System.out.printf("How many adjective are there?"); 
int q = in.nextInt(); 
String adjectiveArray[]=new String[q]; 

System.out.printf("How many adjective are there?"); 
int r = in.nextInt(); 
String objectArray[]=new String[r]; 

System.out.printf("How many adverbs are there?"); 
int p = in.nextInt(); 
String adverbArray[]=new String[p]; 

    Scanner input = new Scanner(System.in); // aanother scanner for more input 

for(int subjectCount=0;subjectCount<t;subjectCount++){ 
subjectArray[subjectCount]= input.next(); 
    int tempSub= subjectCount+1; 
System.out.printf("Subject "+tempSub); 
// Assigned users value to array 
} 
int subjectIndex; 
Random randt = new Random(); 
subjectIndex = randt.nextInt(+1);//for loop 
// random word 
//Stores and finds random subject 
//System.out.println(subjectArray[subjectIndex]); 

for(int verbCount=0;verbCount<o;verbCount++){ 
verbArray[verbCount]= input.next(); // Assigned users value to array 
    int tempVerb= verbCount+1; 
    System.out.printf("verb "+tempVerb); 
    } 
int verbIndex; 
Random rando = new Random(); 
verbIndex = rando.nextInt(o); // random word 
// 


for(int adjectiveCount=0;adjectiveCount<r;adjectiveCount++){ 
adjectiveArray[adjectiveCount]= input.next(); // Assigned users value to array 
    int tempAdj= adjectiveCount+1; 
    System.out.printf("Adjective " + tempAdj); 
} 
int adjectiveIndex; 
Random randq = new Random(); 
adjectiveIndex = randq.nextInt(r); // random word 
// 

for(int objectCount=0;objectCount<q;objectCount++){ 
objectArray[objectCount]= input.next(); // Assigned users value to array 
    int tempObj= objectCount+1; 
    System.out.printf("Object "+ tempObj); 
} 
int objectIndex; 
Random randr = new Random(); 
objectIndex = randr.nextInt(q); // random word 
// 

for(int adverbCount=0;adverbCount<p;adverbCount++){ 
adverbArray[adverbCount]= input.next(); // Assigned users value to array 
    int tempAd= adverbCount+1; 
    System.out.printf("Adverb "+tempAd); 
} 
int adverbIndex; 
Random randp = new Random(); 
adverbIndex = randp.nextInt(p); 

input1 = (subjectIndex + " " + verbIndex +" "+ adjectiveIndex +" "+ objectIndex +" "+ adverbIndex); // adds all items togather for loop 
while(loop = true){ 
System.out.println(input1); 
break; 
} 

}} 

Answer 1

問題：

在幾個你寫while (something = true) {... break; }地方。顯然，你的意思是讓循環運行到something等於true，但是你卻發現循環是永遠運行的，所以在最後加上了break;的快速修復。

問題是，=是轉讓;平等測試是==。你就OK了，如果你寫了：

while (something == true) { ... } 

你也可以寫在較短的版本

while (something) { ... } 

因爲something是boolean，因此已經擁有了自己true或false價值的價值，您無需針對true（或false）進行測試，以獲取您的循環需要的boolean條件。編寫something == true或something == false實際上並不是錯誤的，只是比必要的更冗長。

建議：

您在隨機挑選的話，形成一個句子的基本邏輯基本上是好的，但它需要從用戶獲得這些詞列表的過程中解開，特別是如果你想使用最後一個循環來生成多個隨機句子。

獲得了不同詞類的隨機索引後，您需要從詞列表中提取相應的詞以實際構建句子;現在你只是打印出由索引本身組成的「句子」。

您不需要多個Random對象;它不是特別有害，只是額外的不必要的代碼行和幾個浪費的CPU週期。