jagguy Posted April 8, 2007 Share Posted April 8, 2007 I get a warning problem and i can't see anythong wrong with the code. Also this fails on my livesite but not on my pc. Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 204 in /home/jagguy/public_html/zen/includes/languages/english/html_includes/custom/define_myspecials.php on line 77 Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 204 in /home/jagguy/public_html/zen/includes/languages/english/html_includes/custom/define_myspecials.php on line 78 Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 204 in /home/jagguy/public_html/zen/includes/languages/english/html_includes/custom/define_myspecials.php on line 79 $sql2="select a.categories_id ,parent_id ,categories_name from products_to_categories a ,categories b ,categories_description c where a.categories_id= b.categories_id and b.categories_status=1 and c.categories_id= b.categories_id and a.products_id='$pid'"; $result2 = mysql_query($sql2) or die("Invalid query: " . mysql_error()); $parid= mysql_result($result2, 0, 'parent_id'); //line77 $catid= mysql_result($result2, 0, 'categories_id');//line 78 etc $catname= mysql_result($result2, 0, 'categories_name'); Link to comment https://forums.phpfreaks.com/topic/46121-warning-problem/ Share on other sites More sharing options...
PC Nerd Posted April 8, 2007 Share Posted April 8, 2007 i dont know much about the _result, but look at actually getting the array mysql_fetch_array, and looping through that i think thats the more popular way of doing it, and i know whats how id do it, but i dont know for certain, i dont use _result. good luck Link to comment https://forums.phpfreaks.com/topic/46121-warning-problem/#findComment-224127 Share on other sites More sharing options...
per1os Posted April 8, 2007 Share Posted April 8, 2007 $sql2="select a.categories_id ,parent_id ,categories_name from products_to_categories a ,categories b ,categories_description c where a.categories_id= b.categories_id and b.categories_status=1 and c.categories_id= b.categories_id and a.products_id='$pid'"; $result2 = mysql_query($sql2) or die("Invalid query: " . mysql_error()); $parid= mysql_result($result2, 1, 'parent_id'); //line77 $catid= mysql_result($result2, 1, 'categories_id');//line 78 etc $catname= mysql_result($result2, 1, 'categories_name'); Try that. If not that than this will work: $sql2="select a.categories_id ,parent_id ,categories_name from products_to_categories a ,categories b ,categories_description c where a.categories_id= b.categories_id and b.categories_status=1 and c.categories_id= b.categories_id and a.products_id='$pid'"; $result2 = mysql_query($sql2) or die("Invalid query: " . mysql_error()); $row = mysql_fetch_array($result2); $parid = $row['parent_id']; $catid = $row['categories_id']; $catname = $row['categories_name']; Now this will work for 1 row, which it seems like you are doing if you want more than one than you would need to put that into a while loop. Hope that works out for ya. Link to comment https://forums.phpfreaks.com/topic/46121-warning-problem/#findComment-224231 Share on other sites More sharing options...
jagguy Posted April 9, 2007 Author Share Posted April 9, 2007 thanks Link to comment https://forums.phpfreaks.com/topic/46121-warning-problem/#findComment-224851 Share on other sites More sharing options...
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