Confused Guy Posted September 7, 2014 Share Posted September 7, 2014 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, Link to comment Share on other sites More sharing options...
HartleySan Posted September 7, 2014 Share Posted September 7, 2014 Hello and welcome to the forums. Could you please explain in more detail what is incorrectly occurring? Thanks. Link to comment Share on other sites More sharing options...
Confused Guy Posted September 7, 2014 Author Share Posted September 7, 2014 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 Link to comment Share on other sites More sharing options...
HartleySan Posted September 7, 2014 Share Posted September 7, 2014 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. 1 Link to comment Share on other sites More sharing options...
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