Here's PHP RatingScript

DivaX007

got this script to work, but i after it runs, i get this error. Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c: ..\ rate.php on line 19

<?php

// database variables

$server="localhost"; $user=XXXX"; $pass="XXXXX"; $db="Gallery";
$connection=mysql_connect($server,$user,$pass);

$q="SELECT * FROM photogallery WHERE id='$id' ";

$result= mysql_db_query($db, $q, $connection);

if(!$result)
{
$error_number = mysql_errno();
$error_msg = mysql_error();
echo "MySQL error $error_number: $error_msg";
}

while ($row=mysql_fetch_array($result))
{
$id=$row["id"];
$title=$row["title"];
$photo=$row["photo"];
$Num_Votes=$row["Num_Votes"];
$Votes =$row["Votes"];
$Rating=$row["Rating"];

$new_Votes=$Num_Votes+1;
$Votes=$Votes+$Rate;
$Rating=round(($Votes/$new_Votes),1);

$q="update photogallery set Num_Votes='$new_Votes', Votes='$Votes', Rating='$Rating' where id='$id'";

$result= mysql_db_query($db, $q, $connection) or die
("Could not execute query : $q." . mysql_error());

if ($result) {
echo "Thank you. The article has rating = $Rating after your vote.";
}

}

ednark

try and do a [man]print_r/man on $result to see what is in it...
print_r will use '\n' endline characters

so either wrap your print_r in <pre></pre> tags or do a

header('Content-type: text/plain');

at the beginning of your script so the print_r becomes readable

bobthedog

Well - sounds like your sql query is screwed up then. I would guess either the table "photogallery" doesn't exist (or the user 'XXXX' doesnt have select permission on it), or maybe your database "Gallery" doesn't exist (or the user 'XXXX' doesnt have access to it),or maybe the column 'id' does not exist

Try running this query manually through MySQL and see what happens. You don't mention whether your error string is being dumped out... cos these could provide you with the answer ya looking for!

BTW, reading the manual (which is a favourite pastime of mine) brings this up: -

Note: This function has been deprecated since PHP 4.0.6. Do not use this function. Use mysql_select_db() and mysql_query() instead.

Weedpacket

...and, since if there is an error, it doesn't make much sense to try and continue running, put a die() or something else to gracefully terminate script execution if $result is false.

Weedpacket

...and put the results of your two queries in separate variables, otherwise the second one will end up stomping all over the other.

bobthedog

... and putting your code in PHP ... /PHP tags, cos it makes it a hell of a lot easier to read!

milind24

Hello ,

Check first mysql query , if it give some problem i.e fetches no record don't go down and execute other SQL statemnets use if else condition , It will work .

DivaX007

i changed this:

$result= mysql_db_query($db, $q, $connection) or die

("Could not execute query : $q." . mysql_error());

to this:

$result2 = mysql_db_query($db, $q, $connection) or die

("Could not execute query : $q." . mysql_error());

if ($result2) {

echo "Thank you. The article has rating = $Rating after your vote.";

}

and i worked! i was trying to use the same variable for two different outcomes. thanks

😃