如何将值从PHP表单插入另一个SQL查询

Question

I have created a PHP form with radio buttons so on submit it will produce a roomCode based on the one I have selected. I am struggling to show the other SQL statement once the submit button has been pressed.

PHP:

<?php 
    if( isset( $_POST['submit'] ) )
    {
        if(isset($_POST['proj_check']))
            $proj_check = $_POST['proj_check'];
        else
            $proj_check = "N";

        $numeroOption= $_POST['numero'];
        $roomtype= $_POST['roomtype'];
        $selectOption = $_POST['parkname'];
        $query = "SELECT * FROM `ROOMS` 
            WHERE `Capacity` < '$numeroOption' 
            AND `Park` LIKE '$selectOption%'
            AND `dataProjector` LIKE '$proj_check%' 
            AND `Whiteboard` LIKE '$white_check%' 
            AND `OHP` LIKE '$ohp_check%' 
            AND `WheelchairAccess` LIKE '$wheel_check%' 
            AND `lectureCapture` LIKE '$cap_check%' 
            AND `Style` LIKE '$roomtype%'"; 
        $result = mysql_query($query);
        if ($result == FALSE) die ("could not execute statement $query<br />");
                echo "<form action='' method='post'>";
        echo "<table>"; 

        while($row = mysql_fetch_array($result)){                           
                echo "<tr><td>" . $row['roomCode'] . "</td></tr>";
                echo "<td><input type='radio' name='radioSelect' value= '". $row['roomCode']."'></td>";
                }
        echo "<input type='submit' name='ttroom' id='ttroom' name='ttroom'>";
        echo "</tr>";
        echo "</table>"; 
        echo "</form>";
    }

    if( isset( $_POST['ttroom'] ) )
    {
            $roomcode = $_POST['ttroom'];
            $try = "SELECT * FROM 'ROOMBOOKING' WHERE 'roomCode' = '$roomcode';";
            $ttres = mysql_query($try);
            if ($ttres == FALSE) die ("could not execute statement $try<br />");

                echo "<table>"; 

            while($ttrow = mysql_fetch_array($ttres)){                          
                echo "<tr><td>" . $ttrow['roomCode'] . "</td>";
            }
            echo "</tr>";
            echo "</table>"; 
    }               
    mysql_close();              
?>

Answer 1

Here's an example of something that I assume you're trying to do:

Website:

http://www.w3schools.com/php/php_form_complete.asp

What the above link does is basically creates a form, sets some form validation and asks the user to complete the form. Once the user has completed the form and submitted the form, it echoes the results entered. Now I'm assuming that is what you're looking for. And if you're worried about communicating with the database, you could basically place all the form field names into individual variables and then run an insert query at the same time. So you can echo based on your insert query. Now I noticed in your code, you used a table... forms, tables.. the concept is the same. Give it a go, let me know what you get!

Hope this helps,

Sohail