java - jtextarea 视口(viewport)中的第一行和最后一行

Question

我正在寻找一个函数，它从 jtextarea 给出视口(viewport)起始行和视口(viewport)结束行。下面的代码工作正常。但是当 jtextarea 中的行数太大时，比如 10,000 行，光标的响应变得很慢。我缩小了导致它的范围，它是，

 startLine = getRow(topLeft, editorTextArea) - 1; //editorTextArea is jtextarea name
 endLine = getRow(bottomRight, editorTextArea);

我在每个 keyPressEvent 上调用 startAndEndLine()

有人可以建议我一个更好的代码吗？

private void startAndEndLine() {

    Rectangle r = editorTextArea.getVisibleRect();
    Point topLeft = new Point(r.x, r.y);
    Point bottomRight = new Point(r.x + r.width, r.y + r.height);

    try {
        startLine = getRow(topLeft, editorTextArea) - 1;
        endLine = getRow(bottomRight, editorTextArea);
    } catch (Exception ex) {
       // System.out.println(ex);
    }        
}


 public int getViewToModelPos(Point p, JTextComponent editor) {
    int pos = 0;
    try {
        pos = editor.viewToModel(p);
    } catch (Exception ex) {
    }
    return pos;
 }


public int getRow(Point point, JTextComponent editor) {
    int pos = getViewToModelPos(point, editor);
    int rn = (pos == 0) ? 1 : 0;
    try {
        int offs = pos;
        while (offs > 0) {
            offs = Utilities.getRowStart(editor, offs) - 1;
            rn++;
        }
    } catch (BadLocationException e) {
        System.out.println(e);
    }
    return rn;
}

Answer 1

这是基于 JigarJoshi 在这个问题 Java: column number and line number of cursor's current position 中提出的解决方案...您一定会喜欢这个网站；)

protected int getLineNumber(int modelPos) throws BadLocationException {

    return textArea.getLineOfOffset(modelPos) + 1;

}

Rectangle viewRect = scrollPane.getViewport().getViewRect();

Point startPoint = viewRect.getLocation();
int pos = textArea.viewToModel(startPoint);

try {

    int startLine = getLineNumber(pos);

    Point endPoint = startPoint;
    endPoint.y += viewRect.height;

    pos = textArea.viewToModel(endPoint);
    int endLine = getLineNumber(pos);

    System.out.println(startLine + " - " + endLine);

} catch (BadLocationException exp) {
}

这并不完全准确，但为您提供了一个起点。