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??
Saving PHP output as a file
I wondered if anyone knows how to save the rendered output of a PHP script as a file?I have a script that creates dynamic PDF documents, but want to save it as a file, rather than render it to the
php call servlet
I have done a php backup application .So there is a form that user pick some files to zip and download.Because of a problem with greek characters I make a servet (java) to create the zip.But I want
Check if another session of the page is running?
I want to keep people from opening multiple tabs, or sessions of my Facebook app. Is it possible to check if a session is currently running of my app, like maybe using a cookie or something?
rand() function
just a general question guys a girls, is the rand() function 100% random or is it based on time?
Hits this week counter
I have a counter on my site that tracks hits, IP's, etc. into a mySQL database. It also includes the date in the entry. The code I'm using, that's not working, to try and retrieve the amount of hits
download directory onto C drive
I am attempting (if this is possible) to write a routine to automatically dump the contents of a directory on the server into a directory on the local C drive.I started out like this :Code:
To add a field on the screen XK02.
Hi All,
Error in Configure System Landscape Directory phase
Hi
retrieving images from mysql database using php
So I've been trying to figure out how to store images in a mysql database, and as far as i can tell the images are stored but getting them out seems to be the problem. when i try to go to the page on
Load file in PHP
I have 2 files. The first is a PHP generated XML file that's dependent by 2 inputs. The second is a PHP file that grabs the content of the XML file.So I have