[RESOLVED] The UPDATE function

jerrylouise

Where am i going wrong

I want the following to update all the feilds in the database with the new information relating to the ID. but it keeps erroring out.

$id = $_GET['id'];
$poster = $_POST['poster'];
$author = $_POST['author'];
$title = $_POST['title'];
$body = $_POST['body'];

$server = mysql_connect($host, $username, $password);
$db = mysql_select_db($database, $server); 
$query = 'UPDATE `storys` SET body=$body WHERE id=$id';
mysql_query($query) or die('Error, query failed');
if(!$sql) {
echo "There was an error, please try again.";
print mysql_error();
} 
else 
echo "uploaded";
?>

NogDog

Any values which are not numeric must be quoted in your query. So, if column "body" is a string while "id" is an integer, then your query would be:

$query = "UPDATE `storys` SET `body`='$body' WHERE `id`=$id";

jerrylouise

errors out with

Notice: Undefined variable: sql in /home/fhlinux168/d/dev.cossycomforts.com/user/htdocs/edit/story_edit_uploader.php on line 28
There was an error, please try again.

//grab data from form
$id = $_GET['id'];
$poster = $_POST['poster'];
$author = $_POST['author'];
$title = $_POST['title'];
$body = $_POST['body'];

$server = mysql_connect($host, $username, $password);
$db = mysql_select_db($database, $server); 
$query = "UPDATE `storys` SET `body`='$body' WHERE `id`=$id";
mysql_query($query) or die('Error, query failed');
if(!$sql) {
echo "There was an error, please try again.";
print mysql_error();
} 
else 
echo "uploaded";
?>

alimadzi

That has nothing to do with your query. You're trying to test a variable called $sql that doesn't exist.

jerrylouise

Ok i have made it all work perfict now

//grab data from form
$id = $_GET['id'];
$author = $_POST['author'];
$title = $_POST['title'];
$body = $_POST['body'];

$server = mysql_connect($host, $username, $password);
$db = mysql_select_db($database, $server); 
$query = "UPDATE `storys` SET body = '$body', title = '$title', author = '$author' WHERE id = '$id'";
mysql_query($query) or die('Error, query failed');
if(!$query) {
echo "There was an error, please try again.";
print mysql_error();
} 
else 
echo "uploaded";
?>

NogDog

alimadzi wrote:
That has nothing to do with your query. You're trying to test a variable called $sql that doesn't exist.

Heh--I didn't see that first time through. Of course, giving us the actual error message helps. 🙂

Presumably you want:

$sql = mysql_query($query)......

alimadzi

OK, so it works but not perfectly. Where you say "if not $query", that will always be true because $query contains your actual query. You probably want to change that to something like this...

if ( !mysql_query( $query ) ) {
    echo "There was an error, please try again.";
    echo mysql_error( );
}

bpat1434

To further alimadzi's comments, this:

mysql_query($query) or die('Error, query failed');
if(!$query) {
echo "There was an error, please try again.";
print mysql_error();
}

should really be more like this:

$result = mysql_query($query) or die('Error, query failed');
if(mysql_affected_rows() < 1) {
echo "There was an error, please try again.";
print mysql_error();
}