Can i get a little help please.

[Phpfr3ak]

<?php



include("server.php");


$sql = "UPDATE players SET points = points + 5 WHERE is_active = 1 ORDER BY score DESC LIMIT 50";

$sql = "INSERT INTO credit_logs(player_id,Description,amount,end_balance) VALUES ($playerID,top 50 finish,5,5)";


mysql_query($sql) or die(mysql_error());

echo "Prizes Paid"



?>

Unsure as to why im getting; 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 'top 50 finish,5,5)' at line 1

 

sorry my php isnt too great any help would be great thanks.

 

[AyKay47]

1. by setting 2 values to $sql, you are overiding the first sql string..

2. you are receving this error because you have not wrapped your values in quotes.. the only time in sql that you do not need to wrap a value in quotes is if its an integer..and even then its not a bad idea to..

 

include("server.php");
$sql = "UPDATE players SET points = points + 5 WHERE is_active = 1 ORDER BY score DESC LIMIT 50";
$sql1 = "INSERT INTO credit_logs(player_id,Description,amount,end_balance) VALUES ('$playerID','top 50 finish',5,5)";
$query = mysql_query($sql) or die(mysql_error());
$query1 = mysql_query($sql1) or die(mysql_error());
if($query && $query1){ //if both queries are successful
echo "Prizes Paid";
}

[Phpfr3ak]

Thanks for that i now get how i went about it wrong, will know in future most appreciated

 

still getting an odd message:

Parse error: syntax error, unexpected T_VARIABLE in C:\Program Files\EasyPHP-5.3.3\www\public_html\crons\payprizes.php on line 4

 

Any clue? i'll try and work it out for myself if not just a tad lost atm

[AyKay47]

this normally means that you forgot a line terminator " ; " somewhere in your code.. check the end of line 3