Need help updating data in MYSQL database from PHP page

[DiscoTrio]

I used to be good at this but I changed servers and everything is different...  Heres my code so far:

 

<?php
$rated=$_REQUEST['rated'];
echo $rated;
$rating=$_REQUEST['rating'];
echo $rating;

// Make a MySQL Connection
mysql_connect("localhost", "********", "********") or die(mysql_error());
mysql_select_db("*********") or die(mysql_error());

$result = mysql_query("SELECT * FROM main WHERE username = '$rated'")
or die(mysql_error());  

$row = mysql_fetch_array( $result );

$votes = $db_field['$rating'];
$newvotes = $votes + 1;

echo $newvotes;

mysql_query("UPDATE main SET $rating = '$newvotes'
WHERE username = '$rated'");


?>

 

Whats going on here is the colomb that I want to update comes as a variable $rated (That works) and then the database selects the  row to update with $username (That works) and gets the variable $newvotes by taking the original value of the data its about to update and add 1 to it (That works)  Then it updates the field to $newvotes....

 

I don't know why the update won't go through... there are no errors....

[Twitch]

maybe try without ' 

"UPDATE main SET $rating = $newvotes
WHERE username = $rated"

 

or try

"UPDATE main SET $rating ='".$newvotes."'
WHERE username = '".$rated."'"

 

is the database column really named $rating or is it simply rating?

[DiscoTrio]

It really is called that.

[DiscoTrio]

Hmm putting the query in directly says this:

 

The following errors were reported:You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1 ='1' WHERE username ='DiscoTrio'' at line 1

[Twitch]

Does the query work with the data hard coded?

"UPDATE main SET $rating = '5'
WHERE username = '1'"

 

Not sure what your actual data is, I just picked some random numbers. 

 

[DiscoTrio]

Nope, thats what I just did and got that error.

[gevans]

You should be able to do this with a single uery, something like this (please note the comments on the first few lines, I assumed $_POST data;

 

<?php

/**
* $_REQUEST contains all od the data from $_POST, $_GET and $_COOKIE
*
* If you know where the data is coming from you should use the correct array
* I've guessed $_POST
*/
$rated=$_POST['rated'];
echo $rated;
$rating=$_POST['rating'];
echo $rating;

// Make a MySQL Connection
mysql_connect("localhost", "********", "********") or die(mysql_error());
mysql_select_db("*********") or die(mysql_error());


$result = mysql_query("UPDATE `main` SET `rating` = (`rating` + 1)
    WHERE `username` = '$rated'");

if($result)
{
    echo 'good';
}
else
{
    echo 'bad';
}
?>

 

I assumed that the field in the database that you're updating is `rating`. If this is called something different replace both instances of `rating` with the correct name.

[DiscoTrio]

Im already not liking my new hosting company, now I can't get a sing query to work, Gevans, your query makes it return "Your query did not return any results. (0.0004 seconds)"

[Twitch]

gevans, I assumed it was called 'rating' as well, but he said it really was called $rating.  I wasn't even aware that you could use $ in the name, but I just tried it and mysql let me...haha

[gevans]

Im already not liking my new hosting company, now I can't get a sing query to work, Gevans, your query makes it return "Your query did not return any results. (0.0004 seconds)"

 

The server you're using should make no difference if you're still using php and mysql (as I'm guessing you were before). Also an update query shouldn't attempt to return results, It would just return true or false.

 

If you use phpmyadming try using the following query in the sql tab;

 

UPDATE `main` SET `rating` = `rating` + 1 WHERE `username` = 'PUT A USERNAME IN HERE'

 

Just replace 'PUT USERNAME IN HERE' with a username you know is in the db.

[DiscoTrio]

Its called the value that $rating returns, which is 1,2,3,4  I have that established and it works.

 

And this server has SQL Buddy, I use PHPMYADMIN before...

[gevans]

So let me just get it clear. This page finds two varialbes (I'm assuming via $_POST) '$rated' and '$rating'. Then you want to update the db called 'main'. Update the 'rating' field by '$rating' where the username is equal to '$rated'??

 

If this is correct then a little change and you should have no problem.

[DiscoTrio]

yes

[gevans]

<?php
$rated = $_POST['rated'];
echo $rated;
$rating = (int) $_POST['rating'];
echo $rating;

// Make a MySQL Connection
mysql_connect("localhost", "********", "********") or die(mysql_error());
mysql_select_db("*********") or die(mysql_error());


$result = mysql_query("UPDATE `main` SET `rating` = `rating` + $rating
    WHERE `username` = '$rated'");

if($result)
    echo 'good';
else
    echo 'bad';
?>

 

Now if everything works you should get the username and rating printed, followed by either 'good' or 'bad'.

[DiscoTrio]

DiscoTrio3bad

 

For some reason the query won't go through the database... 

[gevans]

So there's an error in the SQL somewhere, I'm guessing you have errors turned off, add;

 

ini_set('display_errors', 1);
ini_set('error_reporting', E_ALL);

 

To the top of the script, just after the opening php tag and reply with your results!

[DiscoTrio]

Here is another reason my host sucks:

 

Warning: ini_set() has been disabled for security reasons in /www/zzl.org/r/o/b/robloxfashion/htdocs/voting.php on line 10

 

Warning: ini_set() has been disabled for security reasons in /www/zzl.org/r/o/b/robloxfashion/htdocs/voting.php on line 11