Question

尝试在Java中创建ArrayList时遇到问题，但更具体地说是在尝试add()时遇到问题。我在people.add(joe);行...

上遇到语法错误

Error: misplaced construct: VariableDeclaratorId expected after this token.
    at people.add(joe);
                  ^

我的理解是ArrayList对于我的目的而言会比数组更好，所以我的问题是，是不是，我的语法错在哪里？

这是我的代码......

import java.util.ArrayList;

public class Person {
    static String name;
    static double age;
    static double height;
    static double weight;

    Person(String name, double age, double height, double weight){
        Person.name = name;
        Person.age = age;
        Person.height = height;
        Person.weight = weight;
    }

    Person joe = new Person("Joe", 30, 70, 180);
    ArrayList<Person> people = new ArrayList<Person>();
    people.add(joe);
}

Answer 1

static String name;      
static double age;
static double height;
static double weight;

为什么这些变量定义为静态？

看起来你在Person类中这样做了。在类中执行它是可以的（可以这样做），但是如果要创建Person对象的ArrayList则没有多大意义。

这里的要点是必须在实际的方法或构造函数或其他东西（实际的代码块）中完成。同样，我不完全确定Person类型的ArrayList在Person类中是多么有用。

import java.util.ArrayList;

public class Person 
{                   // Don't use static here unless you want all of your Person 
                    // objects to have the same data
   String name;
   double age;
   double height;
   double weight;

   public Person(String name, double age, double height, double weight)
   {
      this.name = name;       // Must refer to instance variables that have
      this.age = age;         // the same name as constructor parameters
      this.height = height;    // with the "this" keyword. Can't use 
      this.weight = weight;    // Classname.variable with non-static variables
   }

}

public AnotherClass 
{
   public void someMethod()
   {
      Person joe = new Person("Joe", 30, 70, 180);
      ArrayList<Person> people = new ArrayList<Person>();
      people.add(joe);
      Person steve = new Person("Steve", 28, 70, 170);
      people.add(steve);            // Now Steve and Joe are two separate objects 
                                    // that have their own instance variables
                                    // (non-static)
   }
}

Answer 2

将该代码放在主方法中：

 public class Person {

     public static void main(String[] args ) {
         Person joe = new Person("Joe", 30, 70, 180);
         ArrayList<Person> people = new ArrayList<Person>();
         people.add(joe);
     }
 }

Answer 3

在一些方法或块中写下这些喜欢::

import java.util.ArrayList;

public class Person 
{
   static String name;
   static double age;
   static double height;
   static double weight;

  Person(String name, double age, double height, double weight)
  {
    Person.name = name;
    Person.age = age;
    Person.height = height;
    Person.weight = weight;
  }

  public static void main(String args[])
  {
    Person joe = new Person("Joe", 30, 70, 180);
    ArrayList<Person> people = new ArrayList<Person>();
    people.add(joe);
  }
}

Answer 4

ArrayList add操作应该在方法块中完成（可能在main中），就像其他人建议的那样。

public class Person  {
  static String name;
  static double age;
  static double height;
  static double weight;

  Person(String name, double age, double height, double weight) {
    Person.name = name;
    Person.age = age;
    Person.height = height;
    Person.weight = weight;
  }
}

public static void main(String[] args) {
  Person joe = new Person("Joe", 30, 70, 180);
  ArrayList<Person> people = new ArrayList<Person>();
  people.add(joe);
}

Answer 5

将代码（在构造函数之后）放在方法中，然后在其他地方调用该方法。您的字段不得为此目的而静态，因为所有实例都将共享它。

Answer 6

首先，您的ArrayList变量确实应该位于实例级别，或static。声明如下：

private ArrayList<Person> people;

或：

static ArrayList<Person> people;

其次，您需要某种功能来执行操作。

public static void addPerson(Person p) {
    people.add(p);
}

第三，你需要调用它。您可以通过以下方式执行此操作：

Person.addPerson(new Person("Joe", 30, 70, 180));

在main()的正文中，或与您的程序执行相关的某个地方。

Answer 7

把花括号括起来 people.add（JOE）; 代码将编译：

{people.add(joe);}

为什么呢？在初始化块上留下练习。

Java ArrayList如何添加类

7 个答案: