小编enj*_*jal的帖子

我想改变jfreechart的背景颜色.它以灰色显示,我想要一个白色背景.我试过chart.setBackgroundPaint(Color.WHITE); 但它并没有向我展示白色背景.我有以下代码显示该图

chart.setBackgroundPaint(Color.WHITE); 

我该如何显示白色背景？

如何在sql查询中传递java字符串变量.我已经完成了所有的JDBC连接.

我的sql数据库查询是

sql = "Select * 
       from production AS cust 
       INNER JOIN location AS comp 
       ON cust.location_id = comp.location_id 
       where comp.name = locationnames AND crop_id =1";

它不起作用.但是,如果我执行以下代码,它的工作原理

sql = "Select * 
       from production AS cust 
       INNER JOIN location AS comp 
       ON cust.location_id = comp.location_id 
       where comp.name = "\taplejung"\  
       AND crop_id =1";

现在告诉我如何将变量名称传递给sql查询来执行此操作.Jst告诉我如何将变量locationnames传递给comp.name.

我的完整java函数如下所示:locationCombo表示在组合框中选择的项目.CropCombo也表示相同......

public void displayYearwise() throws SQLException, ClassNotFoundException{

       //jComboBox4.setSelectedItem("Crops");
        //DefaultCategoryDataset dataset = new DefaultCategoryDataset();
         XYSeriesCollection dataset = new XYSeriesCollection();
         XYSeries series = new XYSeries("production");
         XYSeries series1 = new XYSeries("scat"); …

我以表格形式表示数据没有问题.但是,当我单击删除按钮时,我希望我的数据库的$ id即row [0]传递到另一个页面delete_user_post.php,在那里我可以使用$ _REQUEST捕获$ id的这个值.当我回显$ id时,它会显示其相关的id.但是我无法在下一页传递那个值.我做错了什么？`

<tbody>
        <?php while($row = mysql_fetch_array($result)){
        ?>

          <tr>  
            <td><?php echo $row[0] ?></td>  
            <td><?php echo $row[1] ?></td>  
            <td><?php echo $row[2] ?></td>
            <td>
                <div class="control-group">
                <label class="control-label"></label>
                <div class="controls"> 

                <?php $id = $row[0];
                echo $id;
                ?>

                <a href="delete_user_post.php?roomid=$id" class="btn btn-success" type="button">Delete</a> 
                </td>   



          </div>
          </tr>  
          <?php } ?>
</tbody>

我的user_delete_post.php看起来像这样:

<?php
      include ('include/DB_Functions.php');      
      $id =$_REQUEST['roomid'];     
      // sending query
      mysql_query("DELETE FROM room_tb WHERE roomID=$id") or die(mysql_error());
?>

如何使用java中的jfreechart在mysql数据库表中绘制数据的散点图.我用过swing库.任何链接都会有所帮助.我搜索谷歌但无法找到理解解决方案.如果你有代码,请提供给我.实际上我做了barchart并使用jfreechart绘制它.我用于条形图的代码就在这里.这里display3函数显示条形图.如何修改它以显示散点图？

public void display3() throws SQLException, ClassNotFoundException{

    DefaultCategoryDataset dataset = new DefaultCategoryDataset();

    String JDBC_DRIVER="com.mysql.jdbc.Driver";
    String DB_URL="jdbc:mysql://localhost/data2";
    Connection conn;
    Statement stmt;
    String USER = "root";
    String PASS = "";
    try{
        Class.forName(JDBC_DRIVER);
        conn=DriverManager.getConnection(DB_URL,USER,PASS);
        System.out.println("Creating statement...");
        stmt = conn.createStatement();

        String sql;
        sql="SELECT * FROM `production` WHERE crop_id = 1 AND location_id = 1";
        ResultSet rs=stmt.executeQuery(sql);

        while (rs.next()){
            //String student = rs.getString("studentname");
            String yeartext = rs.getString("year_of_production");
            //double value = Double.parseDouble(text);
            String productiontext = rs.getString("production_amount");
            double production = Double.parseDouble(productiontext);
            Integer year = Integer.parseInt(yeartext);

            dataset.setValue(production, "production", …