big-dog1965 Posted May 20, 2009 Share Posted May 20, 2009 Can someone show me what the error in this is. Warning: mysql_numrows(): supplied argument is not a valid MySQL result resource n /public_html/venues/venues.php on line 23 This is line 23 $num=mysql_numrows($result); NO Vemues Schedule $query = 'DELETE FROM Venues WHERE DATE_SUB(CURDATE(),INTERVAL 1 DAY) >= `Date1`'; mysql_query($query); $query= 'SELECT venues_field_1, venues_field_2, venues_field_3, venues_field_4, venues_field_5, venues_field_6, venues_field_7, venues_field_8. venues_field_9, venues_field_10, venues_field_11, venues_field_12, venues_field_13, venues_field_14, venues_field_15 FROM Venues ORDER BY venues_field_1 DESC LIMIT 0, 100'; $result=mysql_query($query); $num=mysql_numrows($result); if ($num==0) { echo "<b>NO Vemues Schedule</b><br>"; }else{ echo "<b>Where We Are</b><br>"; } mysql_close(); $i=0; while ($i < $num) { $venues_field_1=mysql_result($result,$i,"venues_field_1"); $venues_field_2=mysql_result($result,$i,"venues_field_2"); $venues_field_3=mysql_result($result,$i,"venues_field_3"); $venues_field_4=mysql_result($result,$i,"venues_field_4"); $venues_field_5=mysql_result($result,$i,"venues_field_5"); $venues_field_6=mysql_result($result,$i,"venues_field_6"); $venues_field_7=mysql_result($result,$i,"venues_field_7"); $venues_field_8=mysql_result($result,$i,"venues_field_8"); $venues_field_9=mysql_result($result,$i,"venues_field_9"); $venues_field_10=mysql_result($result,$i,"venues_field_10"); $venues_field_11=mysql_result($result,$i,"venues_field_11"); $venues_field_12=mysql_result($result,$i,"venues_field_12"); $venues_field_13=mysql_result($result,$i,"venues_field_13"); $venues_field_14=mysql_result($result,$i,"venues_field_14"); $venues_field_15=mysql_result($result,$i,"venues_field_15"); echo "<b>$venues_field_1<br>$venues_field_2 - $venues_field_3</b><br>$venues_field_4<br>$venues_field_5<br>$venues_field_6<br>$venues_field_7<br>$venues_field_8<br>$venues_field_9<br>$venues_field_10<br>$venues_field_11<br>$venues_field_12<br>$venues_field_13<br>$venues_field_14<br>$venues_field_15<br>"; $i++; } Quote Link to comment https://forums.phpfreaks.com/topic/158938-mysql_numrows-error/ Share on other sites More sharing options...
Maq Posted May 20, 2009 Share Posted May 20, 2009 More than likely your query is failing. Change this line to this and post the output: $result=mysql_query($query) or die(mysql_error()); (I think you "mysql_numrows" and "mysql_num_rows" are the same, but the second one is more popular IMO.) Quote Link to comment https://forums.phpfreaks.com/topic/158938-mysql_numrows-error/#findComment-838259 Share on other sites More sharing options...
Masna Posted May 20, 2009 Share Posted May 20, 2009 Can someone show me what the error in this is. Warning: mysql_numrows(): supplied argument is not a valid MySQL result resource n /public_html/venues/venues.php on line 23 This is line 23 $num=mysql_numrows($result); NO Vemues Schedule $query = 'DELETE FROM Venues WHERE DATE_SUB(CURDATE(),INTERVAL 1 DAY) >= `Date1`'; mysql_query($query); $query= 'SELECT venues_field_1, venues_field_2, venues_field_3, venues_field_4, venues_field_5, venues_field_6, venues_field_7, venues_field_8. venues_field_9, venues_field_10, venues_field_11, venues_field_12, venues_field_13, venues_field_14, venues_field_15 FROM Venues ORDER BY venues_field_1 DESC LIMIT 0, 100'; $result=mysql_query($query); $num=mysql_numrows($result); if ($num==0) { echo "<b>NO Vemues Schedule</b><br>"; }else{ echo "<b>Where We Are</b><br>"; } mysql_close(); $i=0; while ($i < $num) { $venues_field_1=mysql_result($result,$i,"venues_field_1"); $venues_field_2=mysql_result($result,$i,"venues_field_2"); $venues_field_3=mysql_result($result,$i,"venues_field_3"); $venues_field_4=mysql_result($result,$i,"venues_field_4"); $venues_field_5=mysql_result($result,$i,"venues_field_5"); $venues_field_6=mysql_result($result,$i,"venues_field_6"); $venues_field_7=mysql_result($result,$i,"venues_field_7"); $venues_field_8=mysql_result($result,$i,"venues_field_8"); $venues_field_9=mysql_result($result,$i,"venues_field_9"); $venues_field_10=mysql_result($result,$i,"venues_field_10"); $venues_field_11=mysql_result($result,$i,"venues_field_11"); $venues_field_12=mysql_result($result,$i,"venues_field_12"); $venues_field_13=mysql_result($result,$i,"venues_field_13"); $venues_field_14=mysql_result($result,$i,"venues_field_14"); $venues_field_15=mysql_result($result,$i,"venues_field_15"); echo "<b>$venues_field_1<br>$venues_field_2 - $venues_field_3</b><br>$venues_field_4<br>$venues_field_5<br>$venues_field_6<br>$venues_field_7<br>$venues_field_8<br>$venues_field_9<br>$venues_field_10<br>$venues_field_11<br>$venues_field_12<br>$venues_field_13<br>$venues_field_14<br>$venues_field_15<br>"; $i++; } You have a very basic (and relatively obvious) syntax error in your query: a period where a comma should be. Use this: $query= 'SELECT venues_field_1, venues_field_2, venues_field_3, venues_field_4, venues_field_5, venues_field_6, venues_field_7, venues_field_8, venues_field_9, venues_field_10, venues_field_11, venues_field_12, venues_field_13, venues_field_14, venues_field_15 FROM Venues ORDER BY venues_field_1 DESC LIMIT 0, 100'; Also, the function is mysql_num_rows(), not mysql_numrows (although, apparently the latter works also). Quote Link to comment https://forums.phpfreaks.com/topic/158938-mysql_numrows-error/#findComment-838260 Share on other sites More sharing options...
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