supplied argument is not a valid MySQL result resource
Posted on 2006-10-20
I know this question has been asked a number of times but I cannot seem to get the solution from looking at the previous responses.
I have a php page that doing a number of calculations to display a league table.
I know that the connection to the database is fine as the previous queries on the same page are all working ok.
The code that is causing the problem is as follows.
88.$mysql = "SELECT ap.player_id,ap.position
89. FROM mfhc_apps ap
90. LEFT JOIN mfhc_matches ma ON ma.match_id = ap.match_id
91. WHERE ap.player_id IN (".$myin.") AND ma.comp_id IN (1,2,3,4) AND ma.oppscore = 0";
92. $myresult = mysql_query($mysql) or die(mysql_error());
94. //Set up loop to work out clean sheets
95. while ($temp=mysql_fetch_array($myresult))
...some other code
The $mysql generates the following query
SELECT ap.player_id,ap.position FROM mfhc_apps ap LEFT JOIN mfhc_matches ma ON ma.match_id = ap.match_id WHERE ap.player_id IN (27,30,98,36,26,56,5,33,91,87,68) AND ma.comp_id IN (1,2,3,4) AND ma.oppscore = 0
Which if entered directly into MYPHPADMIN produces a valid result.
The error message I am getting is as follows
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/martonhockey_co_uk/tests/fantasyleague.php on line 95
This has been bugging me for ages know am getting a bit sick, please help