Where am I going wrong

Posted on 16th Feb 2014 07:03 pm by admin

Been trying to work this out for hours

I have two tables called 'Genres' and 'Films'. Genres contains two columns 'Genre_id' and 'Genre_Name'. Films contains a load of columns, the most important being 'Title', 'Year', Director', 'Genres'. Now im trying to create a page where the user can edit a film in the 'Films' table. I want this form to already be populated with the existing values which the user can then delete and enter new ones. Heres the code:

Code:
<?
include 'connect.php';
$FilmID= $_GET['id'];
$result = mysql_query("SELECT Films.*, Genres.Genre_id, Genres.Genre_Name FROM Films INNER JOIN Genres ON Films.Genres = Genres.Genre_id WHERE Films.Film_id = $FilmID" ,$linkme) or die ('Error: '.mysql_error ());

$row = mysql_fetch_array($result);
$current_Genre = $row['Genre_Name'];
$current_Genre_id = $row['Genre_id'];

?>
<form method="post" action="update2.php?id=<? echo $FilmID ?>">
<table>
<tr>
<td>Title</td>
<td><input name = "Title" value="<? echo $row['Title'] ?>" type="text" /></td>
</tr>
<tr>
<td>Year</td>
<td><input name = "Year" value="<? echo $row['Year'] ?>" type="text" /></td>
</tr>
<tr>
<td>Director</td>
<td><input name = "Director" value="<? echo $row['Director'] ?>" type="text" /></td>
</tr>
<tr><td>
Genre:</td><td> <select name="Genre">
<option value = "<?php echo $current_Genre_id ?>"> <? echo $current_Genre ?> </option>
<?
$result = mysql_query("SELECT * FROM Genres",$linkme) or die ('Error: '.mysql_error ());
while($row = mysql_fetch_array($result));
{ ?>
<option value = "<?php echo $row['Genre_id'] ?>"> <? echo $row['Genre_Name'] ?> </option>
<? } ?>
</select></td>
</tr>
</table>
<input name="Update" type="submit" value="Update" />
</form>
<?
include 'close.php';
?>
<a href = "../admin.php">Back</a>


and heres the second page that actually changes the values:

Code:
<?
include 'connect.php';
$Title = $_REQUEST["Title"];
$Year = $_REQUEST["Year"];
$Director = $_REQUEST["Director"];
$Genre = $_REQUEST["Genre"];
$FilmID = $_GET['id'];

$result = mysql_query("UPDATE Films SET Title = '$Title' , Year = '$Year', Director ='$Director', Genre ='$Genre' WHERE Film_id = '$FilmID'",$linkme);
include 'close.php';
?>
Record Updated<br />
<a href = "../admin.php">Back</a>


There are no errors, the values just dont show up in the form, Where am I going wrong??

Your Answer:

Login to answer

Other forums

SAP Logon Failed
I tried to login to SAP through MMC.

When i click start and give password.

it

Problem with DB connection
Hello there! I'm new to this forum and I'm new to PHP coding also. I wrote something that doesn't ma

GMail like Chat in ASP.NET
Hi,Can anyone suggest me, how to incorporate GMail like chat in my existing ASP.Net application.I wa

help me fix these syntax errors...
I keep getting multiple syntax errors on this script like this one:

Parse error: syntax error

I have a parse error in this query help..
Code: $query1="INSERT INTO `rating` (`item_name`, `rating`, `ip_address`, `date_rated`) VALUES

Trigger tag in aspx is not coming
hi all ,iam new to ajax, iam using file upload inside the update pannel but right now i want to use

help connecting a form to php and then emailing the form
Ok. So, I made a form, and I need help to where when submitted, I get an email with the submitted de

problem with php mysql query
Hi guy's...

I'm totally lost here..because don't have any idea how to make a query for grab r

Consuming third party Payment Gateway API from Procedure.
Hi All,

First of all i would like to thank all the people of the oracle forum for providi

Displaying an image using echo command
I had done a query on my database, one of the fields being an image reference to a directory where a