java编译错误：找不到符号。

Compile error in java :cannot find symbol.

提问by user2977165

I have this code that is taking a text file and turning it into a string and then separating parts of the string into different elements of an arraylist.

我有这段代码，它正在获取一个文本文件并将其转换为一个字符串，然后将字符串的一部分分成一个数组列表的不同元素。

import java.util.Scanner;
import java.io.*;
import java.util.ArrayList;

public class Grocery{

    public Grocery(){

        File inFile = new File ("lists.txt");
        Scanner input = new Scanner (inFile);
        String grocery;
        {
             grocery = input.nextLine();
        }
    }

    public void makeSmallerLists(){
        String listLine;
        String line;
        ArrayList<String> smallList = new ArrayList<String>();
        while(input.hasNextLine()){
            line = input.nextLine;
            if(line.equals("<END>")){
                smallList.add(listLine);
            } else{
                listLine = listLine + "\n" + line;
            }
        }
    }
}

However when I try to compile this it gives me two errors:

但是，当我尝试编译它时，它给了我两个错误：

javac Message.java Message.java:31: cannot find symbol symbol : variable input location: class Message while(input.hasNextLine()){ ^ Message.java:32: cannot find symbol symbol : variable input location: class Message line = input.nextLine; ^

javac Message.java Message.java:31: 找不到符号符号: 变量输入位置: class Message while(input.hasNextLine()){ ^ Message.java:32: 找不到符号符号: 变量输入位置: class Message line = input.nextLine; ^

How do I fix this? I really don't know what's wrong.

我该如何解决？我真的不知道怎么了。

I fixed that and now my error says $ javac Message.java Message.java:34: cannot find symbol symbol : variable nextLine location: class java.util.Scanner line = input.nextLine; ^

我解决了这个问题，现在我的错误是 $ javac Message.java Message.java:34: cannot find symbol symbol : variable nextLine location: class java.util.Scanner line = input.nextLine; ^

           ^

Now what is wrong?

现在出了什么问题？

采纳答案by Suresh Atta

 Scanner input = new Scanner (inFile);

inputis local to the constructor, you cannot access outside of it, and you are trying to access in makeSmallerLists()method. Make it as a instance member, So that it available through out the classother than staticcontext.

input对构造函数来说是本地的，你不能在它之外访问，并且你试图在makeSmallerLists()方法中访问 。将其作为实例成员，以便它在上下文之外的class其他内容中可用static。

public class Grocery{

  Scanner input;

and in constructor

并在构造函数中

public Grocery(){

            File inFile = new File ("lists.txt");
             input = new Scanner (inFile);

回答by SudoRahul

That is because the Scanner object inputhas been declared inside your constructor(local scope of the constructor) and thus its not visible in your makeSmallerLists(). You need to declare it as an instance variable so that it would be accessible in all the methods of the class.

那是因为 Scanner 对象input已在您的构造函数（构造函数的本地范围）中声明，因此它在您的makeSmallerLists(). 您需要将其声明为实例变量，以便在类的所有方法中都可以访问它。

public class Grocery {
    Scanner input; // declared here, as an instance variable
    public Grocery(){
        File inFile = new File ("lists.txt");
        input = new Scanner (inFile);  // initialized here
        ...
    }
    ...
    public void makeSmallerLists() {
        ...
        while(input.hasNextLine()) { // accessible everywhere within the class
        ...
    }
}

回答by Masudul

You have variable scope problem. You can't get access a field outside of scope. Declare Scanner as globally, outside of costructor.

您有可变范围问题。您无法访问范围之外的字段。在 costructor 之外将 Scanner 声明为全局。

 public class Grocery{
    Scanner input = null;// Declare Scanner here.

    public Grocery(){
        .....
        input=new Scanner (inFile);
    }

Also append method bracket ().

还附加方法括号()。

      public void makeSmallerLists(){
       ......
        while(input.hasNextLine()){
        line = input.nextLine();// Append () after method. 
       .....
      }

回答by Maroun

One solution is to have a class memberof type Scanner:

一种解决方案是拥有一个类型为的类成员Scanner：

private Scanner input;

And in the constructor, construct it:

在构造函数中，构造它：

public class Grocery { 
   private Scanner input;

   public Grocery() {
      ...
      ...
      input = new Scanner (inFile);
   }
   ...
}

Now inputis not limited to the scopeof the constructor, it'll be accessible through the whole class.

现在input不限于构造函数的范围，它可以通过整个类访问。

---

---

Consider this example:

考虑这个例子：

public class Example {
    int n = 0;  //n known everywhere in the class

    if(n == 0) {
       int n1 = 1;
       if(n1 == 1) {
          int n2 = 2;
          //n1 is known here
       }
       //n2 is not known here
    } 
    //n1 is not known here
}

回答by Vinayak Pahalwan

inputis not accesible outside the constructor...it's declared insisde the constructor

input在构造函数之外不可访问……它在构造函数内部声明