jQuery - Save to SQL via PHP
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by Kenny Bones
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Published on 2010-05-06T11:34:13Z
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2010/05/06
11:38 UTC
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This is probably easy for you guys, but I can't understand it. I want to save the filename of an image to it's own row in the SQL base.
Basically, I log on to the site where I have my own userID. And each user has its own column for background images. And the user can choose his own image if he wants to. So basically, when the user clicks on the image he wants, a jquery click event occurs and an ajax call is made to a php file which is supposed to take care of the actual update. The row for each user always exist so there's only an update of the data that's necessary.
First, I collect the filename of the css property 'background-image' and split it so I get only the filename. I then store that filename in a variable I call 'filename' which is then passed on to this jQuery snippet:
$.ajax({
url: 'save_to_db.php',
data: filename,
dataType:'Text',
type: 'POST',
success: function(data) {
// Just for testing purposes.
alert('Background changed to: ' + data);
}
});
And this is the php:
<?php
require("dbconnect.php")
?>
<?php
$uploadstring = ($_POST['filename']);
mysql_query("UPDATE brukere SET brukerBakgrunn = $uploadstring WHERE brukerID=" .$_SESSION['id'] ."";
mysql_close();
?>
Basically, each user has their own ID and this is called 'brukerID' The table everything is in is called 'brukere' and the column I'm supposed to update is the one called 'brukerBakgrunn'
When I just run the javascript snippet, I get this message box in return where it says:
Background changed to:
Parse error: syntax error, unexpected ';' in /var/www/clients/client2/web8/web/save_to_db.php on line 8
I actualle get this messagebox twice, not sure why. Line 8 in 'save_to_db.php' is this one:
mysql_query("UPDATE brukere SET brukerBakgrunn = $uploadstring WHERE brukerID=" .$_SESSION['id'] ."";
Not sure if you need to see db_connect.php as well. I can add that later if you need to see it. So what am I missing here?
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