我有一个用于过滤JTable中的数据（多个或没有）的方法，这是代码：
public void filter() {
if(Main.is_loading)
return;
RowFilter<Object, Object> serviceFilter = null;
if (!multiple) {
String filterText = tfield.getText();
if (filterText.length() > 2)
serviceFilter = RowFilter.regexFilter("(?iu)" + filterText);
} else {
List<RowFilter<Object, Object>> filters = new ArrayList<RowFilter<Object, Object>>(2);
for (int i = 0; i < tfields.size(); i++) {
String filterText = "";
JTextField cur = tfields.get(i);
filterText = cur.getText();
if (cur.getText().length() > 2) {
filters.add(RowFilter.regexFilter("(?iu)" + filterText, i));
}
}
serviceFilter = RowFilter.andFilter(filters);
}
sorter.setRowFilter(serviceFilter);
table.setRowSorter(sorter);
}
如何更改找到的符号的颜色？我需要标记匹配的符号。
最佳答案
首先过滤表模型以查找匹配的单元格。然后，对于所有匹配的单元格，强制重新绘制单元格视图（如果使用了AbstractTableModel调用fireTableCellUpdated(int,int)）。另外，您还需要改进单元格渲染器。当您使用文本过滤器和单元格文本匹配时，只需显示具有其他颜色的匹配文本即可。
class Renderer extends DefaultTableCellRenderer {
@Override
public Component getTableCellRendererComponent(JTable table, Object value, boolean isSelected, boolean hasFocus, int row, int column) {
if (isFilterUsed() && isCellMatched(row, column)) {
String stringValue = value.toString();
String filter = getFilterValue();
int start = stringValue.indexOf(filter);
int end = start + filter.length();
String result = String.format("<html>%s<font color='red'>%s</font>%s</html>",
stringValue.substring(0, start),
stringValue.substring(start, end),
stringValue.substring(end));
return super.getTableCellRendererComponent(table, result, isSelected, hasFocus, row, column);
} else {
return super.getTableCellRendererComponent(table, value, isSelected, hasFocus, row, column);
}
}
}