Question

我正在做关于java I / O数据的功课，问题是我不允许使用对象序列化从二进制文件加载数据。

以下是作业要求：

持久性，将对象写入文件并从文件中读取对象（文本格式）
• All objects should be written to a single file in a text format
• All objects should be read form the same file
• You should use JFileChooser

我有3个课程：Unit，Assessment和Task。

Assessment类是抽象的IndividualAssessment＆amp; GroupAssessment是具体的子类。

Unit有Assessment的集合，Assessment有Task的集合。

我可以使用FileWriter将所有数据保存到一个文本文件中，但我不知道如何将文本文件的每一行读入正确的Assessment类。

我的意思是，您如何识别IndividualAssessment或GroupAssessment类的哪一行。

这是我尝试的代码，但它不起作用：

BufferedReader bf = new BufferedReader(file);
While (bf.readLine != null){
    Unit u = new Unit(bf);
    diary.add(u);
    try{
        Assessment a = new IndividualAssessment(bf);
    } catch (IOException ex){
        Assessment a = new GroupAssessment(bf);
            }
    u.add(a);
    Task t = new Task(bf);
    a.add(t);

Answer 1

您应该从存储在String对象中读取的行的内容开始。所以它看起来像这样：

String line = bf.readLine();
while(line != null) {
    // Add code to look at your line to figure out what kind of object it 
    // represents.   For example you could add a one character prefix to each 
    // line when you write it to specify which object it is ('U', 'A' or 'T').
    // Based on that, you can call a constructor of the appropriate object that
    // takes a String as input. Then let the constructor deal with parsing the
    // line for the object it represents. This way you don't end up with some
    // massive parsing routine.

    char code = line.charAt(0);
    if(code == 'T') {
        Task task = new Task();
        task.initFromString(line.sustring(1));
        ... Do something with your task
    }
    else if(code == ...) {
    }

    line = bf.readLine();    // Read the next line
}

定义您的所有对象应实现的一些接口：

public interface TextExportable {
    public char getClassIdentifier();
    public void initFromString(String s);
}

我不确定你的物品是什么样子但是比如说：

public class Task implements TextExportable {
    private String name;

    public Task() {} // For the pseudo-serialization

    public Task(String name) { // For creating the object by hand
        this.name = name;
    }

    public char getClassIdentifier() {
        return 'T';
    }

    public String toString() {
        return getClassIdentifier()+name;
    }

    public void initFromString(String line) {
        this.name = line;
        // Here, you would need extra parsing if you have more than one attribute
        // and dissect the line
    }
}

Answer 2

在构造函数中使用BufferedReader作为参数是一种非常规的方法来解决这个问题。您实例中的BufferedReader是交付模型数据的工具。并且您的模型数据应该与您的用例有关，而不是关于如何将其存储在磁盘上，因此它应该足够抽象。所以相反，我会在每个模型类上都有一个默认构造函数，然后可能是一个构造函数，可用于对模型类的所有/某些字段进行初始化。

接近文件界面的一种方法是创建文件读取器/写入器工具，可以从/向文件加载/保存数据。如果类层次结构的顶部是Unit，那么我会这样：

Unit loadFromFile(String filename) {
   // read unit, assessments etc in a format you choose, build the model and return it
   ...
}

void saveToFile(String filename) {
   // write you model to a file
   ...
}

如果我是你，我会使用JSON作为我的数据格式。它可读，易于调试。

我希望这会有所帮助。

Answer 3

除了接受的答案之外，我想补充一点，一种干净的解析方法是遵循interpreter pattern

所以假设bnf的语法如下：

<Unit>        ::= <Name>[<Assessments>]
<Assessments> ::= <Assessment> | <Assessment> <Assessments>
<Assessment>  ::= <Name>:<Type>[<Tasks>]
<Tasks>       ::= <Task> | <Task>,<Tasks>
<Name>        ::= <String>
<Type>        ::= Group | Ind

你可能会写这样的东西来解析它 - 我把它留作练习来编写所使用的Expression类。

try {
    BufferedReader br = new BufferedReader(new FileReader("File.txt"));
    try {
        String unit;
        while ((unit = br.readLine()) != null) {
            String unitName = unit.substring(0, unit.indexOf('['));
            String assessments = unit.substring(unit.indexOf('[') + 1,
                            unit.lastIndexOf(']'));
            List<AssessmentExpression> aes = new ArrayList<AssessmentExpression>();
            while (!assessments.equals("")) {
                String assessmentName = assessments.substring(0,
                                assessments.indexOf(':'));
                String assessmentType = assessments.substring(
                                assessments.indexOf(':') + 1,
                                assessments.indexOf('['));
                String tasks = assessments.substring(
                                assessments.indexOf('[') + 1,
                                assessments.indexOf(']'));
                List<String> taskList = Arrays.asList(tasks.split(","));
                aes.add(new AssessmentExpression(assessmentName,
                                assessmentType, taskList));
                assessments = assessments.substring(assessments.indexOf(']')+1);
            }

            Unit u = new UnitExpression(unitName, aes).interpret();
        }
    } finally {
        br.close();
    }
} catch (IOException e) {
    e.printStackTrace();
}

从java中的文本文件中读取对象数据

3 个答案: