Why Won't My Database Update?

[ryanfilard]

I am trying to make a rating system and this script is not working.

 

$ratingget = $_REQUEST['star2'];
mysql_select_db($database_news, $news);
$query_ratingsys = "UPDATE news SET votes = votes +1 AND rating = rating +'$ratingget' WHERE id = '$id'";
$ratingsys = mysql_query($query_ratingsys, $news) or die(mysql_error());

[MasterACE14]

are there any errors showing up? and is $_REQUEST['star2'] set?

[ryanfilard]

No here is my form that submits.

 

<form method="post" action="view.php?id=<?PHP echo $_REQUEST['id']; ?>"> 
                              <input name="star2" type="radio" class="star" value="1"/>
<input name="star2" type="radio" class="star" value="2"/>
<input name="star2" type="radio" class="star" value="3"/>
<input name="star2" type="radio" class="star" value="4"/>
<input name="star2" type="radio" class="star" value="5"/> 
<input type="hidden" name="formcheck" value="rating">
<input name="rate" type="submit" id="rate" value="Rate"> 
</form>

[BigTime]

Hi Ryan

 

# check for errors

if (!$ratingsys) { echo("ERROR: " . mysql_error() . "\n$query_ratingsys\n");

 

$query_ratingsys = "UPDATE news SET votes = votes +1 AND rating = rating +'$ratingget' WHERE id = '$id'";

 

vote and rating you are trying to add to are not variables.

 

Also, use a comma where the AND is

 

Before I processed the update request I would define $votes, then add the +1 as $newvotes

define $rating then add $ratingget as $newrating

 

Then UPDATE news SET votes='$newvote',rating='$newrating' WHERE id = '$id'";

 

but Im a not very efficient hack and there is probably a smarter way

[Buddski]

Is MySQL throwing any errors?

My guess is you are trying to "add" a string to a, presumably, integer database column.

Try something like this.

$query_ratingsys = "UPDATE `news` SET `votes` = (`votes`+1) AND `rating` = (`rating`+$ratingget) WHERE `id` = '$id'";

[ryanfilard]

I used this

$query_ratingsys = "UPDATE news SET votes=votes+'$newvote',rating=rating+'$ratingget' WHERE id = '$id'";
[code]
It works now thank you.