1

以下代码将从文本字段中获取文本并在 JTable 中搜索它。它仅显示文本的第一次出现。我也需要连续发生。所以，请指导我如何实现这一目标。提前致谢。

private void search8()
{
    String target8 = sear8.getText();
    for(int row = 0; row < table8.getRowCount(); row++)
        for(int col = 0; col < table8.getColumnCount(); col++)
        {
            String next8 = (String)table8.getValueAt(row, col);
            if(next8.equals(target8))
            {
                showSearchResults(row, col);
                return;
            }
        }
}

更新：

private void showSearchResults(int row, int col)
{
    CustomRenderer renderer = (CustomRenderer)table8.getCellRenderer(row, col);
    renderer.setTargetCell(row, col);
    Rectangle r8 = table8.getCellRect(row, col, false);
    table8.scrollRectToVisible(r8);
    table8.repaint();
}

class CustomRenderer implements TableCellRenderer
{
    public CustomRenderer()
    {
       label = new JLabel();
       label.setHorizontalAlignment(JLabel.CENTER);
       label.setOpaque(true);
       targetRow = -1;
       targetCol = -1;
    }

    public Component getTableCellRendererComponent(JTable table,
    Object value,boolean isSelected,boolean hasFocus,int row, int column)
    {
       if(isSelected)
       {
           label.setBackground(table.getSelectionBackground());
           label.setForeground(table.getSelectionForeground());
       }
       else
       {
           label.setBackground(table.getBackground());
           label.setForeground(table.getForeground());
       }
       if(row == targetRow && column == targetCol)
       {
           label.setBackground(new Color(176,196,222));
           //label.setBorder(BorderFactory.createLineBorder(Color.red));
           label.setFont(table.getFont().deriveFont(Font.BOLD));
       }
       else
       {
           label.setBorder(null);
           label.setFont(table.getFont());
       }
       label.setText((String)value);
       return label;
    }

    public void setTargetCell(int row, int col)
    {
       targetRow = row;
       targetCol = col;
    }
}
4

3 回答 3

3

您在找到 target8 的第一个匹配项后立即返回。

最好建立一个匹配字符串列表并传递给您的方法showSearchResults。

于 2012-07-27T13:25:14.230 回答
3

有两种方法可以实现这一点：

  • 正如 Reimeus 建议的那样，执行一次解析以查找特定搜索关键字的所有出现并将它们的位置存储在列表中。然后调用你的showSearchResults()并遍历列表。

  • 第二个选项是将您上次扫描的位置存储在某个地方。然后当用户按下下一步时，从这个位置开始搜索，而不是再次 (0,0)。

我个人更喜欢第一个选项，因为我不必重复扫描此表。此外，此列表将帮助我轻松实现“上一个”和“下一个”功能

编辑：这是实现此目的的一种方法（请注意，您必须根据自己的要求对其进行自定义，这只是为了帮助您入门）：

private void search8() {
    ArrayList<String> resultList = new ArrayList<String>();

    String target8 = sear8.getText();
    for (int row = 0; row < table8.getRowCount(); row++) {
        for (int col = 0; col < table8.getColumnCount(); col++) {
            String next8 = (String) table8.getValueAt(row, col);
            if (next8.contains(target8)) {
                resultList.add(row + "," + col);
            }
        }
    }

    System.out.println(sarr8.length);

    String[] tokens;
    for (String result : resultList) {
        try {
            tokens = result.split("[,]");
            showSearchResults(tokens[0], tokens[1]);
        } finally {
            tokens = null;
        }
    }

    /**
     * Your remaining part
     */
}
于 2012-07-27T14:16:53.753 回答
1 
// I'd, personally, make this protected as you may wish to change the how the search 
// is performed in the future.
protected void search8() {

    // You've assumed that there are only ever 40 elements
    // while you've allowed for a variable number of search positions
    // You would need (at least) (rowCount * colCount) * 2 elements to be
    // safe.  This is a little ridiculous considering that there might
    // only be 1 reuslt in the table
    // int[] sarr8 = new int[40]; <-- Don't really want to do this

    // Instead, we should use a dynamic array instead
    // The ArrayList is a Collection implementation backed by an array
    // but it has the means to grow (and shrink) to meet the capacity requirements
    List<Point> slist8 = new ArrayList<Point>(25); // <-- you could change the initial value as you see fit
    int i = 0;
    String target8 = sear8.getText();
    for (int row = 0; row < table8.getRowCount(); row++) {
        for (int col = 0; col < table8.getColumnCount(); col++) {
            String next8 = (String) table8.getValueAt(row, col);
            if (next8.contains(target8)) {
                // Okay, this kinda cheating, but we want to store col/row or x/y
                // cell coordinates.  You could make your own class "Cell" class,
                // but for what we want, this is exactly the same
                Point cell = new Point(col. row);
                //sarr8[i] = row;
                //sarr8[i + 1] = col;
                //i = i + 2;
                slist8.add(cell);
            }
        }
    }

    //System.out.println(sarr8.length);
    System.out.println(slist8.size());

    //for (int j = 0; j < sarr8.length; j += 2) {
    //    showSearchResults(sarr8[j], sarr8[j + 1]);
    //    return;
    //}

    // Now, personally, I'd pass in the whole result set to the "showSearchResults"
    // method, because, IMHO, that's the methods domain of responsibility, ours was
    // to simply find the results.

    showSearchResults(slist8);

    // At this point, the showSearchResults method can determine how it wants to display
    // the search results

}

@Sujay 在他的回答中也证明了这种方法

更新

for (Point p : slist8) {
    showSearchResults(p.x, p.y);
}

别的

private void showSearchResults(List<Point> results)
{

    for (Point p : results) 
    {
        int col = p.x;
        int row = p.y;
        CustomRenderer renderer = (CustomRenderer)table8.getCellRenderer(row, col);
        renderer.setTargetCell(row, col);
        Rectangle r8 = table8.getCellRect(row, col, false);
        table8.scrollRectToVisible(r8);
    }
    table8.repaint();
}
于 2012-07-27T22:09:18.403 回答