java 在java中创建一个存储字符串和整数的数组
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Creating an array that stores strings and integers in java
提问by Billy
I would like to create an array that stores the names (String) and values (integer) of company stocks, but I don't know how to go about it.
我想创建一个数组来存储公司股票的名称（字符串）和值（整数），但我不知道如何去做。
回答by Stephen C
An Object[]can hold both Stringand Integerobjects. Here's a simple example:
AnObject[]可以同时容纳String和Integer对象。这是一个简单的例子：
Object[] mixed = new Object[2];
mixed[0] = "Hi Mum";
mixed[1] = Integer.valueOf(42);
...
String message = (String) mixed[0];
Integer answer = (Integer) mixed[1];
However, if you put use an Object[]like this, you will typically need to use instanceofand / or type casts when accessing the elements.
但是，如果Object[]像这样使用 use instanceof，则在访问元素时通常需要使用和/或类型转换。
Any design that routinely involves instanceofand/or type casts needs to be treated with suspicion. In most cases, there is a better (more object-oriented, more efficient, less fragile) way of achieving the same ends.
任何经常涉及instanceof和/或类型转换的设计都需要怀疑。在大多数情况下，有一种更好的（更面向对象、更高效、更不脆弱）的方式来实现相同的目标。
In your particular use-case, it sounds like what you really need is a mapping object that maps from String(names) to Integer(numbers of stocks). And the nice thing about Java is that there are existing library classes that provide this functionality; e.g. the HashMap<K,V>class, with Stringas the key type and Integeras the value type.
在您的特定用例中，听起来您真正需要的是一个从String（名称）映射到Integer（股票数量）的映射对象。 Java 的好处是现有的库类提供了这个功能；例如HashMap<K,V>类，String作为键类型和Integer作为值类型。
Another possibility might be an array, Listor Setof some custom or generic pair class. These have different semantic properties to Maptypes.
另一种可能性可能是一个数组，List或者Set一些自定义或通用对类。它们对Map类型具有不同的语义属性。
回答by Billy
You can either use array declaration or array literal (but only when you declare and affect the variable right away, array literals cannot be used for re-assigning an array).
您可以使用数组声明或数组文字（但仅当您立即声明并影响变量时，数组文字不能用于重新分配数组）。
For primitive types:
对于原始类型：
int[] myIntArray = new int[3];
int[] myIntArray = {1,2,3};
int[] myIntArray = new int[]{1,2,3};
For classes, for example String, it's the same:
对于类，例如 String，它是相同的：
String[] myStringArray = new String[3];
String[] myStringArray = {"a","b","c"};
String[] myStringArray = new String[]{"a","b","c"};
Extra Info:If you are knew to coding in Java here are some tutorials:http://www.youtube.com/watch?v=Cfd9DOnuF9w
额外信息：如果您了解 Java 编码，这里有一些教程：http: //www.youtube.com/watch?v=Cfd9DOnuF9w
-HAPPY CODING!
- 编码快乐！
回答by Rene M.
You have two choices:
你有两个选择：
- Use an array.
- 使用数组。
public class Value { public String name; public int number; } ... public Value[] values = new Value[10]; ....
public class Value { public String name; public int number; } ... public Value[] values = new Value[10]; ....
- Use a map which has much more comfort, specially you can use the name as key to get a value
- 使用更舒适的地图，特别是您可以使用名称作为键来获取值
.... public Map<String, int> valueMap = new HashMap<String,int>(); valueMap.put("Sample",10); int value = valueMap.get("Sample"); ...
.... public Map<String, int> valueMap = new HashMap<String,int>(); valueMap.put("Sample",10); int value = valueMap.get("Sample"); ...
回答by mohsaied
You can use a Map data structure instead of an array. This is basically a type of Collection that has key-value pairs. Your string name can be used as the key and the value is your integer.
您可以使用 Map 数据结构而不是数组。这基本上是一种具有键值对的 Collection 类型。您的字符串名称可以用作键，值是您的整数。
Map<String,Integer> myMap = new HashMap<String, Integer>;
MyMap.put("someone", 6);
Note that using a HashMap has a speed advantage over an array during lookup. The complexity of HashMap lookup is O(log(n)) while that of an array is O(n).
请注意，在查找期间使用 HashMap 比数组具有速度优势。HashMap 查找的复杂度是 O(log(n)) 而数组的复杂度是 O(n)。
