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java - 在Java中对两个相同名称的文本文件进行排序

转载 作者:行者123 更新时间:2023-12-01 12:40:20 25 4
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我目前正在开发一个 Java 项目,

以下是我迄今为止的编码尝试:

     public class MyZoo
{
// zoo identifier
private String zooId;
// a number used in generating a unique identifier for the next animal to be added to the zoo
private int nextAnimalIdNumber;
// zstorage for the Animal objects
private TreeMap<String, List<Animal>> animals;

/**
* Create an "empty" zoo.
*
* @param zooId an identifier for the zoo, at least three characters long.
*/
public MyZoo(String zooId)
{
this.zooId = zooId.trim().substring(0,3).toUpperCase();
nextAnimalIdNumber = 0;
animals = new TreeMap<String, List<Animal>>();
}

/**
* Returns a unique identifier, for an <tt>Animal</tt> object, based on the
* zoo identifier and the field <tt>nextAnimalIdNumber</tt> which is incremented
* ready for next time the method is called.
*
* @return a unique identifier.
*/
public String allocateId()
{
// increment nextAnimalIdNumber and then construct a six digit string from it
nextAnimalIdNumber++;
String s = Integer.toString(nextAnimalIdNumber);
while ( s.length()<6 )
s = "0" + s;
return zooId + "_" + s;
}

/**
* Adds an animal to the zoo.
*
* @param animal the Animal object to be added.
*/
public void addAnimal(Animal animal)
{
String animalName = animal.getName();
// Is there already an animal with the same name?
if (!animals.containsKey(animalName)){
// If it is not in the map, create a list
animals.put(animalName, new ArrayList<Animal>());
}
// Now add the animal to the list
animals.get(animalName).add(animal);
}

/**
* Reads <tt>Animal</tt> data from a text file and adds them to the zoo. The
* format of the data is specified in the MyZoo coursework assignment.
*
* @param animal the Animal object to be added.
*/
public void readDataFromFile()
{
int noOfAnimalsRead = 0;

// set up an owner for the FileDialog
JFrame jframe = new JFrame();
jframe.setVisible(true);
// use a Filedialog to select a file to read from
FileDialog fDialog = new FileDialog(jframe, "Read from", FileDialog.LOAD);
fDialog.setFile("import001.txt");
fDialog.setDirectory(".");
fDialog.setVisible(true);
String fname = fDialog.getFile();
jframe.dispose();

File inFile = new File(fname);

String fileName = "import002.txt";

// This will reference one line at a time
String line = null;


try {
// FileReader reads text files in the default encoding.
FileReader fileReader =
new FileReader(fileName);

// Always wrap FileReader in BufferedReader.
BufferedReader bufferedReader =
new BufferedReader(fileReader);

while((line = bufferedReader.readLine()) != null) {
System.out.println(line);
}

// Always close files.
bufferedReader.close();
}
catch(FileNotFoundException ex) {
System.out.println(
"Unable to open file '" +
fileName + "'");
}
catch(IOException ex) {
System.out.println(
"Error reading file '"
+ fileName + "'");

}

addAnimal( new Animal("golden eagle", "Eddie", this) ); //
addAnimal( new Animal("tiger", "Tommy", this) );
addAnimal( new Animal("lion", "Leo", this) );
addAnimal( new Animal("parrot", "Polly", this) );
addAnimal( new Animal("cobra", "Collin", this) );

noOfAnimalsRead = 5;

// this next line should be retained
System.out.println("no of animals read from file was " + noOfAnimalsRead + "\n");
}

/**
* Prints out details of all animal in the zoo.
*
*/
public void printAllAnimals()
{

System.out.println("\nDetails for all animals in Zoo " + zooId);
System.out.println( "==================================");

Collection<Animal> c = animals.values();
// The name of the file to open.
String fileName = "import001.txt";

// This will reference one line at a time
String line = null;
for (Map.Entry<String, List<Animal>> animalEntry : animals.entrySet())
{
List<Animal> animalsOfAName = animalEntry.getValue();
for (Animal animal: animalsOfAName){
// output here, change as appropriate, maybe add the type of animal
System.out.println(animal.getName());
}
}
}
}

但是文本文件(Eddie)中有相同的文本名称,如何让第二个Eddie替换第一个Eddie而不丢失字母顺序,在addAnimal()中put()使用名称字段

最佳答案

键在 map 中必须是唯一的,这就是第二个艾迪动物取代第一个的原因。

要在同一个键下保存多个对象,您可以声明一个 Map,它使用 List(或其他适当的 Collection)作为键值对中的值.

TreeMap<String, List<Animal>> animals;

现在添加动物时需要一些逻辑来处理这个问题。

public void addAnimal(Animal animal)
{
String animalName = animal.getName();
// Is there already an animal with the same name?
if (!animals.containsKey(animalName){
// If it is not in the map, create a list
animals.put(animalName, new ArrayList<Animal>();
}
// Now add the animal to the list
animals.get(animalName).add(animal);
}

现在要打印,您需要遍历每个名​​称的动物:

for (Map.Entry<String, List<Animal>> animalEntry : animals.entrySet())
{
List<Animal> animalsOfAName = animalEntry.getValue();
for (Animal animal: animalsOfAName){
// output here, change as appropriate, maybe add the type of animal
System.out.println(animal.getName());
}
}

关于java - 在Java中对两个相同名称的文本文件进行排序,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25168549/

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