Carry values over from forms to another page via PhP.

[Lewis_1]

Hello. I've created 2 pages that i'm currently working on. One is called; Test.php -- which just includes a basic form in HTML.

<Form name ="form1" Method ="POST" Action ="result.php">

Building Name: <INPUT TYPE = "TEXT" Name ="buildingName">
Room Number: <INPUT TYPE = "TEXT" Name ="roomId">
<INPUT TYPE = "Submit" Name = "Submit" VALUE = "Go">

</FORM>

The other being a page (called results.php), which grabs the information from the form above when the submit button is clicked and queries a database that it's linked to.

<?php

            $buildingName = $_POST['buildingName'];
            $roomId = $_POST['roomId'];

            $sql = mysql_query("SELECT * FROM forestcourt WHERE buildingName=$buildingName AND id=$roomId");


            $id2 = 'id';
            $buildingName = 'buildingName';
            $subBuildings = 'subBuildings';
            $imagePath = 'imagePath';
            $description = 'description';


            $rows2 = mysql_fetch_assoc($sql);
            echo 'Name: ' . $rows2[$buildingName] . '<br/>' . 'Room Number: ' . $rows2[$id2] . '<br/>' . 'Sub Buildings: ' . $rows2[$subBuildings] . '<br/>' . 'Description: ' . $rows2[$description] . '<br/>' . 'Location: ' . '<img src="../' . $rows2[$imagePath] . '"/>' . '<br/><br/>';


        ?>

However, this code does work and displays all the relevant information accuratly when I'm only working with ONE text-input field (Room Number: <INPUT TYPE = "TEXT" Name ="roomId">) on the page; test.php.

I've later tried to expand on this to take into account the buildingName too. I have set it up in the exact same way that I did with the roomId and it seems to just spit out an error.

I'm pretty new to PhP and MySQL and as far as I'm aware the SQL;

$sql = mysql_query("SELECT * FROM forestcourt WHERE buildingName=$buildingName AND id=$roomId");

is correct.

If anyone could help me with this matter It would be a great help!

Thanks a bunch,

Lewis.

[phorce]

Hey Lewis, I don't unerstand what you mean exactly.. It's only working with one of your input fields? Do you mean that only one result from the database is showing?

[Lewis_1]

Sorry that I wasn't specific.

The full code above breaks when it's run.

however, when only having 1 text field on the form, it works

(for example;

<Form name ="form1" Method ="POST" Action ="result.php">
Room Number: <INPUT TYPE = "TEXT" Name ="roomId">
<INPUT TYPE = "Submit" Name = "Submit" VALUE = "Go">
</FORM>

this works -- and with anything in reference to buildingName removed from 'results.php')

[phorce]

I cleaned results up:

<?php

            $buildingName = $_POST['buildingName'];
            $roomId = $_POST['roomId'];

            $sql = mysql_query("SELECT * FROM forestcourt WHERE buildingName=$buildingName AND id=$roomId") or die("There was an error" + mysql_error());           

            $num_rows = mysql_num_rows($sql);

            if(num_rows >= 1)
            {
                 $id2 = 'id';
                 $buildingName = 'buildingName';
                 $subBuildings = 'subBuildings';
                 $imagePath = 'imagePath';
                 $description = 'description';

                 while($row2 = mysql_fetch_assoc($sql))
                 {
                     echo 'Name: ' . $rows2[$buildingName] . '<br/>' . 'Room Number: ' . $rows2[$id2] . '<br/>' . 'Sub Buildings: ' . $rows2[$subBuildings] . '<br/>' . 'Description: ' . $rows2[$description] . '<br/>' . 'Location: ' . '<img src="../' . $rows2[$imagePath] . '"/>' . '<br/><br/>';         
                 }



            }

        ?>

Try inserting both text fields and this should give an error, if there is one.. If there isn't one, then I would check whether or not buildingName is actually a field name. Hope this helps

[Lewis_1]

Hello, Phorce. Thanks for the reply.

It seems that what you gave me doesn't seem to give out any information. It doesn't even say that there was an error.

buildingName is a field name in test.php so I can't see it being that.

Thanks,

Lewis.

[phorce]

But.. is buildingName a field name inside your database/table? And contains the data that you are sending through the form? If it isn't, the query will never be true.

<?php

            $buildingName = $_POST['buildingName'];
            $roomId = $_POST['roomId'];

            $sql = mysql_query("SELECT * FROM forestcourt WHERE buildingName=$buildingName AND id=$roomId") or die("There was an error" + mysql_error());           

            $num_rows = mysql_num_rows($sql);

            if(num_rows >= 1)
            {
                 $id2 = 'id';
                 $buildingName = 'buildingName';
                 $subBuildings = 'subBuildings';
                 $imagePath = 'imagePath';
                 $description = 'description';

                 while($row2 = mysql_fetch_assoc($sql))
                 {
                     echo 'Name: ' . $rows2[$buildingName] . '<br/>' . 'Room Number: ' . $rows2[$id2] . '<br/>' . 'Sub Buildings: ' . $rows2[$subBuildings] . '<br/>' . 'Description: ' . $rows2[$description] . '<br/>' . 'Location: ' . '<img src="../' . $rows2[$imagePath] . '"/>' . '<br/><br/>';         
                 }



            }else{
              echo "Cannot get data from table";


        ?>

Sorry, I forgot the else statement.

[Lewis_1]

Ah. Sorry I didn't realise you were referring to the table/db.

Yes, it is a field name within the table 'forestcourt'.

I'm going to speak to my lecturer tomorrow morning about this since I cannot seem to figure it out and it's 3:15am.

Thank you for the help though, Phorce.