无法根据从php中的列表框获取的值来获取mysql记录。

Question

我一直在尝试通过表单在mysql中添加一条记录。我能够做到这一点。添加记录后，获取所有记录的主键以填充列表框。在列表框的单击事件中，我想获取所选记录的完整详细信息。我已经尝试过以下代码。我可以添加记录并填充列表框，但不能根据从列表框中选择的值获取记录。有没有人能给我一些建议。提前谢谢。我的代码如下：

<!--
    File Name - insert_and_drop_down_list.php
    This file performs following actions - 
    (1) Insert record into the database table
    (2) Fetch records from the same table, Populate listbox
    (3) At the click event of the listbox, display the data
-->

<html>
    <head>
        <title>Insert And Drop Down List</title>
        <SCRIPT language=JavaScript>
            function reload(form)
            {
                var e = document.getElementById("lstName");
                var val = e.options[e.selectedIndex].value;
                //var val=form.lstName.options[form.lstName.options.selectedIndex].value;
                self.location='insert_and_drop_down_list.php?param_name=' + val ;
                //window.alert(val);
                document.write(val);
            }
        </script>

    </head>

    <body>
        <form action="insert_and_drop_down_list.php">
            Name : &nbsp;&nbsp;<input type="text" name="txtName"/></br>
            Age : &nbsp;&nbsp;<input type="text" name="txtAge"></br>
            Gender : &nbsp;&nbsp;<input type="text" name="txtGender"/></br>
            Occupation : &nbsp;&nbsp;<input type="text" name="txtOccupation"/><br>
            <input type="submit" name="btnSubmit" value="Sign Up"/><br/>
        </form>
    </body>
</html>


<?php
    $name=isset($_REQUEST['txtName'])?$_REQUEST['txtName']:"";
    $age=isset($_REQUEST['txtAge'])?$_REQUEST['txtAge']:"";
    $gender=isset($_REQUEST['txtGender'])?$_REQUEST['txtGender']:"";
    $occupation=isset($_REQUEST['txtOccupation'])?$_REQUEST['txtOccupation']:"";

    $con = @mysql_connect("localhost","root","");
    $db = mysql_select_db("db_insert_and_drop_down_list");

    if(isset ($_REQUEST['btnSubmit']))
    {
        $sql = "INSERT INTO tbl_employee(col_name,col_age,col_gender,col_occupation)values('$name','$age','$gender','$occupation')";
        $status=mysql_query($sql);

        $sql2 = "SELECT * FROM tbl_employee ORDER BY col_name ASC";
        $rs = mysql_query($sql2);





        echo "<select name='lstName' onchange=\"reload(this.form)\">
        <option value=''>Select one</option>";

        while($row = mysql_fetch_array($rs))
        {
            echo "<option value='$row[col_name]'>$row[col_name]</option>"."<BR>";
        }
        echo "</select>";

        //echo "<input type='submit' name='btnDisplayRecord' value='Display Record'>";
    }   




    //and strlen($param_name) > 0
    $param_name=isset($_REQUEST['param_name'])?$_REQUEST['param_name']:"";
    echo $param_name;
    if(isset($param_name))
    {
        $sql3 = "SELECT * FROM tbl_employee WHERE col_name='$param_name'";
        $rs_emp = mysql_query($sql3);

        echo "pppppppppppppppppppppppppp";

        echo "$param_name";

        echo "nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn";


        while($row_emp = mysql_fetch_array($rs_emp))
        {
            echo "aaaaaaa";

            echo $row_emp['col_name'];

            //echo "<input type='text' name='txtName' value=".$row_emp['col_emp']."/></br>";
        }
    }
    mysql_close();
?>

和SQL

  CREATE TABLE IF NOT EXISTS `tbl_employee` (
  `col_name` varchar(25) NOT NULL,
  `col_age` int(5) NOT NULL,
  `col_gender` varchar(10) NOT NULL,
  `col_occupation` varchar(25) NOT NULL,
  PRIMARY KEY (`col_name`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

--
-- Dumping data for table `tbl_employee`
--

INSERT INTO `tbl_employee` (`col_name`, `col_age`, `col_gender`, `col_occupation`) VALUES
('aa', 11, 'aa', 'aa'),
('aaa', 11, 'aaaaa', 'aaa');

请看一看，给我指路。谢谢。

Answer 1

• 这是目前正在遵循的命名约定，我很高兴我不需要使用这种命名结构。tbl_，col_...如果你是上面所有代码的所有者，请切换到pdo或mysqli。
• 在一遍又一遍地看代码之后，我发现原来的页面并没有重定向，那就是

self.location='insert_and_drop_down_list.php?param_name=‘+ val ;

此行无法重定向到自身并传递参数。由于参数没有传递，所以整个脚本都会失败。我甚至尝试按如下方式传递参数

<?php echo Header('Location:insert_and_ddl.php?param_name=' + val)?>

但是页面仍然没有重定向，也没有传递参数。有什么建议吗？