Trying to correct a PHP problem

ltd

Hello, first post here. I'm trying to figure out what the below errors mean. Can someone pls explain what these mean?

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/nobody/public_html/deltaforceraiders/sigs/statsig.php on line 31

Warning: Cannot modify header information - headers already sent by (output started at /home/nobody/public_html/deltaforceraiders/sigs/statsig.php:31) in /home/nobody/public_html/deltaforceraiders/sigs/statsig.php on line 45
‰PNG  IHDR†KNO²xPLTEÿáw’û«IDATxœíÁ1Â õOm  àÅ¦2"dýIEND®B`‚

Here are the lines from the php file:

Line 31:

while ($i!=1 && $data = mysql_fetch_object($sql))                 {

Line 45:

header("Content-Type: image/png");

Thanks in advance!

bbreslauer

The mysql_fetch_object() error means that it couldn't get the data from the database. A couple possibilities are: 1) your SQL query is malformed (to check this, echo $sql and then try that query in the mysql command line or phpMyAdmin, for example) or 2) you didn't connect to the database first.

The header error is because headers must be the first thing sent to the browser. This could be cleared up after fixing the other error, as that sent an error message to the browser before the header was sent.

ltd

bbreslauer wrote:
The mysql_fetch_object() error means that it couldn't get the data from the database. A couple possibilities are: 1) your SQL query is malformed (to check this, echo $sql and then try that query in the mysql command line or phpMyAdmin, for example) or 2) you didn't connect to the database first.

The header error is because headers must be the first thing sent to the browser. This could be cleared up after fixing the other error, as that sent an error message to the browser before the header was sent.

Thanks for the reply. I tried to run the sql query from phpmyadmin, now I only ran line 31, is that correct or should I have ran the entire script? I got an error when I tried to run it. Also, I don't know what you mean by echo $sql first......Here is the entire script:

<?php
//+++++++++++++++++++++++++++++
//Fill in your Database details
//+++++++++++++++++++++++++++++
$dbhost="localhost"; //mysql host address
$dbuname="xxxxx_xxx";//mysql user name
$dbpass="xxxxxx"; //mysql user pass
$dbname="xxxx_xxxx"; //babstats db name

mysql_connect($dbhost, $dbuname, $dbpass) or die ("SQL server down");
$db=mysql_select_db($dbname) or die ("Unable to select database");


//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
//If you want to edit this for chronos, edit the query below
//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
$sql = mysql_db_query($dbname,"SELECT id, player_name, total_kills, last_played, total_deaths, total_knives, total_headshots, total_games FROM bab_stats where id = $id");

//++++++++++++++++++++++++++++++++++
//filename of the background picture
//++++++++++++++++++++++++++++++++++
$image = "signature.png";
$im = imagecreatefrompng($image);



//++++++++++++++++++++++
//RGB Colour of the Text
//++++++++++++++++++++++
$tc = ImageColorAllocate ($im, 255, 225, 119);
while ($i!=1 && $data = mysql_fetch_object($sql))                 {


//++++++++++++++++++++++
//change yoursite.com to your website url
//++++++++++++++++++++++
ImageString($im, 3, 105, 5, "Player stats for $data->player_name", $tc);
ImageString($im, 2, 105, 20, "Total kills: $data->total_kills | Total deaths: $data->total_deaths", $tc);
ImageString($im, 2, 105, 30, "Total Knives: $data->total_knives | Total Headshots: $data->total_headshots", $tc);
ImageString($im, 2, 105, 43, "      Last Played: $data->last_played", $tc);
ImageString($im, 2, 105, 55, "Total Games: $data->total_games            www.deltaforceraiders.net", $tc);
$i++;
}echo "";
$db=mysql_close ();
header("Content-Type: image/png");
Imagepng($im,'',100);
ImageDestroy ($im);
?>

thanks again!

bbreslauer

You need to just run the SQL query, which is (from your script):

SELECT id, player_name, total_kills, last_played, total_deaths, total_knives, total_headshots, total_games FROM bab_stats where id = $id

However, you do need to change $id to a valid value, and probably put single quotes (') around $id.

The header function must go before anything is output to the page. Two lines above, you have an echo command, which will probably mean that the header will give you an error instead.

In addition, mysql_db_query() is deprecated; you should use mysql_query() instead. (http://us2.php.net/manual/en/function.mysql-db-query.php)

ltd

Thanks for the reply. I think I found the problem:

SELECT id, player_name, total_kills, last_played, total_deaths, total_knives, total_headshots, total_games FROM bab_stats where id = $id

the table bab_stats does not exist, I changed it to DFR_stats, and when I run the query in myphpadmin, I get the player data. Now I dont get the original errors, and can see the resulting .jpg, but the data from the database is not there:

http://www.deltaforceraiders.net/sigs/statsig.php?id=198&?=.jpg

Also, when I tried to change mysql_db_query() to mysql_query() I got a new error on line 17

thanks again

bbreslauer

that's because mysql_query() doesn't take the database name as a parameter.

ltd

so i guess i would need to change something else too? I think I'll just leave it the way it is......this script written by someone else, & is working on multiple other sites.

The only problem I'm having now is when the image is created, the only sql data written is player_name, none of the other data from the other tables are being written. Now if I change player_name to total_kills, the kills do display. So I'm trying to understand why is that line the only one to display the data???

ImageString($im, 3, 105, 5, "Player stats for $data->player_name", $tc);
ImageString($im, 2, 105, 20, "Total kills: $data->total_kills     Total deaths: $data->total_deaths", $tc);
ImageString($im, 2, 105, 30, "Total Knives: $data->total_knives    Total Headshots: $data->total_headshots", $tc);
ImageString($im, 2, 105, 43, "Last Played: $data->last_played", $tc);
ImageString($im, 2, 105, 55, "Total Games: $data->total_games     www.deltaforceraiders.net", $tc);

Thanks again!

ltd

Would anyone know why the $data on this line is working ok.........

ImageString($im, 3, 105, 5, "Player stats for $data->player_name", $tc);

But the $data on these lines don't???

ImageString($im, 2, 105, 20, "Total kills: $data->total_kills     Total deaths: $data->total_deaths", $tc);
ImageString($im, 2, 105, 30, "Total Knives: $data->total_knives    Total Headshots: $data->total_headshots", $tc);
ImageString($im, 2, 105, 43, "Last Played: $data->last_played", $tc);
ImageString($im, 2, 105, 55, "Total Games: $data->total_games     www.deltaforceraiders.net", $tc);

Here is the link to the created .jpg. Also, when I run the query in phpmyadmin, I can see all the $data just fine.

thanks