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Populating A Table Using Values From The Drop Down List


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Hi

Im having a problem populating a table when selecting different options from the drop down list(DDL)

Heres my code:

 

This is for the DDL

 Please select Catergory: <select name="cat">
                            <option value="n" > Choose one </option>
            				<option value="Bellow1600cc" > Below 1600cc </option>
            				<option value="Diesel" > Diesel </option>
            				<option value="1600cc" > 1600cc </option>
                            <option value="1800cc" > 1800cc </option>
            				<option value="2000cc" > 2000cc </option>
            				<option value="ko3Turbo" > ko3Turbo </option>
                            <option value="ko4Turbo" > ko4Turbo </option>
            				<option value="6Cyl" > 6Cyl </option>
            				<option value="6CylTurbo" > 6CylTurbo </option>
                            <option value="8Cyl" > 8Cyl </option>
            				<option value="4CylAftermarket" > 4CylAftermarket </option>
        </select>
      

And this is the php code:


    <?php
        
        $con=mysql_connect("localhost","root","");
        if(!$con)
        {
            die("could not connect".mysql_error());
        }
        
        mysql_select_db("DynoWarsPS",$con);  
        
        $query="SELECT * FROM $_POST['cat']";
		$result=mysql_query($query,$con);
        $num = mysql_num_rows($result);
		
        $title =" ";
        if(isset($_POST['cat']))
        {
			if($_POST["cat"] == 'Bellow1600cc')
        {   
            $title ="Below 1600cc";
        }
        else if($_POST["cat"] == 'Diesel')
        {   
            $title ="Diesel";
        }
        else if($_POST["cat"] == '1600cc')
        {   
            $title ="1600cc";
        }
        else if($_POST["cat"] == '1800cc')
        {   
            $title ="1800cc";
        }
        else if($_POST["cat"] == '2000cc')
        {   
            $title ="2000cc";
        }
        else if($_POST["cat"] == 'ko3Turbo')
        {   
            $title ="ko3Turbo";
        }
        else if($_POST["cat"] == 'ko4Turbo')
        {   
            $title ="ko4Turbo";
        }
		else if($_POST["cat"] == '6Cyl')
        {   
            $title ="6Cyl";
        }
		else if($_POST["cat"] == '6CylTurbo')
        {   
            $title ="6CylTurbo";
        }
        else if($_POST["cat"] == '8Cyl')
        {   
            $title ="8Cyl";
        }
        else if($_POST["cat"] == '4CylAftermarket')
        {   
            $title ="4CylAftermarket";
        }
        
		}
        echo"<h2> $title </h2>";

        echo "<table border=1 align=center>
        <tr>
            <th>Name</th>
            <th>Vehicle</th>
            <th>PowerKW</th>
            <th>TorqueNM</th>
        </tr>            ";
            
            while($Arr=mysql_fetch_array($result))
		    {
			    echo "<tr>";
                    echo "<td>". $Arr['Name']. "</td>";
                    echo "<td>". $Arr['Vehicle']. "</td>";
                    echo "<td>". $Arr['PowerKW']. "</td>";
                    echo "<td>". $Arr['TorqueNM']. "</td>";
                echo "</tr>";
		    }
		    echo "</table>";
		
		 mysql_close($con);
?>


Im not sure why it doesnt work,

    

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The drop down list is populated with the name of different tables in the database.

By selecting a certain value from the drop down list, the table will be populated with the values from the db

$sql="CREATE TABLE Bellow1600cc(
      Name varchar(10),
      Vehicle varchar(20),
      PowerKW int(3),
      TorqueNM int(4)
       )";
       
$sql="CREATE TABLE Diesel(
      Name varchar(10),
      Vehicle varchar(20),
      PowerKW int(3),
      TorqueNM int(4)
       )";
       

all the table attributes are the same, just table names and values are different

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Try echoing out your query to the screen to make sure it's valid.

Also, try running the query directly on the DB to make sure it actually works.

Lastly, I would recommend using mysqli_query instead of mysql_query.

 

If you're still stump after doing all of that, please let us know.

Thanks.

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