解析Java中的缩进文本树

Z时代
2024-01-10
分类：问答

我有一个缩进文件，需要使用java进行解析，我需要某种方式将其放在Section类中，如下所示

root root1 text1 text1.1 text1.2 text2 text2.1 text2.2 root2 text1 text1.1 text1.2 text2 text2.1 text2.2.2

我有上课的地方像缩进的东西

public class Section 
{
    private List<Section> children;
    private String text;
    private int depth;
    public Section(String t)
    {
       text =t;
    }
    public List<Section> getChildren()
    {
        if (children == null)
      {
            children = new ArrayList<Section>();
       }
        return children;
}
public void setChildren(List<Section> newChildren)
{
    if (newChildren == null) {
        children = newChildren;
    } else {
        if (children == null) {
            children = new ArrayList<Section>();
        }
        for (Section child : newChildren) {
            this.addChild(child);
        }
    }
}
public void addChild(Section child)
{
    if (children == null) {
        children = new ArrayList<Section>();
    }
    if (child != null) {
        children.add(child);
    }
}
public String getText()
{
    return text;
}
public void setText(String newText)
{
    text =newText;
}
public String getDepth()
{
    return depth;
}
 public void setDepth(int newDepth)
 {
    depth = newDepth;
 }
}

我需要一些方法来解析文件并将其放在预期的结果中，这是我们的Section对象，如下所示

Section=
Text="Root"
Children
Child1: Text= "root1"
        Child1: "text1"
            Child1="Text 1.1"
            Child2="Text 1.2"
        Child2: "text2"
            Child1="Text 2.1"
            Child2="Text 2.2"
            Children
Child2: Text= "root2" 
        Child1: "text1"
            Child1="Text 1.1"
            Child2="Text 1.2"
        Child2: "text2"
            Child1="Text 2.1"
            Child2="Text 2.2"
Here is some code that I have started
   int indentCount=0;
   while(String text = reader.readline()
   {
   indentCount=countLeadingSpaces(String word);
   //TODO create the section here
   }
public static int countLeadingSpaces(String word)
{
    int length=word.length();
    int count=0;
   for(int i=0;i<length;i++)
   {
       char first = word.charAt(i); 
        if(Character.isWhitespace(first))
        {
            count++;           
        }
        else
        {
            return count;
        }
   }
 return count;
}

回答：

我也添加了一个父指针。也许不用它也可以解析文本，但是父指针使它更容易。首先，您需要具有更多的构造函数：

static final int root_depth = 4; // assuming 4 whitespaces precede the tree root
public Section(String text, int depth) {
    this.text     = text;
    this.depth    = depth;
    this.children = new ArrayList<Section>();
    this.parent   = null;
}
public Section(String text, int depth, Section parent) {
    this.text     = text;
    this.depth    = depth;
    this.children = new ArrayList<Section>();
    this.parent   = parent;
}

然后，当您开始解析文件时，请逐行读取它：

Section prev = null;
for (String line; (line = bufferedReader.readLine()) != null; ) {
    if (prev == null && line begins with root_depth whitespaces) {
        Section root = new Section(text_of_line, root_depth);
        prev = root;
    }
    else {
        int t_depth = no. of whitespaces at the beginning of this line;
        if (t_depth > prev.getDepth())
            // assuming that empty sections are not allowed
            Section t_section = new Section(text_of_line, t_depth, prev);
            prev.addChild(t_section);
        }
        else if (t_depth == prev.getDepth) {
            Section t_section = new Section(text_of_line, t_depth, prev.getParent());
            prev.getParent().addChild(t_section);
        }
        else {
            while (t_depth < prev.getDepth()) {
                prev = prev.getParent();
            }
            // at this point, (t_depth == prev.getDepth()) = true
            Section t_section = new Section(text_of_line, t_depth, prev.getParent());
            prev.getParent().addChild(t_section);
        }
    }
}