我正在构建一个SQL DB GUI。我遇到的问题是,当用户从SHOW TABLES双击表时,它不会更新显示新表结果的表。
这是我的代码:
JTabbedPane tabbedPane = new JTabbedPane();
Statement stat = Gui.getStat();
private JPanel panel;
private String query;
public static void main(String[] args) {
showTables();
}
public void showTables() throws SQLException{
ResultSet showTablesResult = null;
query = "abc";
try {
showTablesResult = stat.executeQuery("SHOW TABLES");
} catch (SQLException e) {
JOptionPane.showMessageDialog(null, "Error connecting to databse");
}
displayResultSet(showTablesResult);
}
public void displayResultSet(ResultSet rs) throws SQLException{
JTable table = new JTable(buildTableModel(rs)){
private static final long serialVersionUID = 1L;
@Override
public boolean isCellEditable(int row, int column) {
return false;
}
};
panel.add(new JScrollPane(table));
table.addMouseListener(new MouseAdapter(){
public void mouseClicked(MouseEvent e){
if(e.getClickCount() == 2){
if(query == "abc"){
//get value in selected cell
String selectedData = null;
int[] selectedRow = table.getSelectedRows();
int[] selectedColumns = table.getSelectedColumns();
for (int i = 0; i < selectedRow.length; i++) {
for (int j = 0; j < selectedColumns.length; j++) {
selectedData = (String) table.getValueAt(selectedRow[i], selectedColumns[j]);
}
}
try {
viewTable(selectedData);
} catch (SQLException e1) {
e1.printStackTrace();
}
}
}
}
});
}
public void viewTable(String tName) throws SQLException{
ResultSet showTablesResult = null;
query = "123";
try {
showTablesResult = stat.executeQuery("SELECT * FROM " +tName);
} catch (SQLException e) {
JOptionPane.showMessageDialog(null, "Error connecting to databse");
}
displayResultSet(showTablesResult);
}
public static DefaultTableModel buildTableModel(ResultSet rs)
throws SQLException {
ResultSetMetaData metaData = rs.getMetaData();
Vector<String> columnNames = new Vector<String>();
int columnCount = metaData.getColumnCount();
for (int column = 1; column <= columnCount; column++) {
columnNames.add(metaData.getColumnName(column));
}
Vector<Vector<Object>> data = new Vector<Vector<Object>>();
while (rs.next()) {
Vector<Object> vector = new Vector<Object>();
for (int columnIndex = 1; columnIndex <= columnCount; columnIndex++) {
vector.add(rs.getObject(columnIndex));
}
data.add(vector);
}
return new DefaultTableModel(data, columnNames);
}
我该如何解决这个问题使其起作用?我认为可能有一个重绘或刷新表之类的函数,但是我找不到它。
谢谢 :)
panel.add(new JScrollPane(table));
默认情况下,组件的大小为(0,0),因此无需绘制任何内容。
从可见的GUI添加(或删除)组件时,您需要调用布局管理器,以便在面板上为组件指定大小/位置。基本代码是:
panel.add(...);
panel.revalidate();
panel.repaint();
但是,当使用逻辑“刷新”表时,这样做要容易得多:
table.setModel( your updated TableModel );
无需每次都重新创建JScrollPane和JTable。
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