starvinmarvin14 Posted December 14, 2011 Share Posted December 14, 2011 I'm trying to select a random id from a table called 'image'. I want the id number to be at the end of a link such as view.php?id=123. Here is my code: $query=mysql_query("SELECT id FROM image WHERE active='yes' ORDER BY RAND() LIMIT 1"); echo <<<OPT <meta http-equiv="REFRESH" content="0;url=http://www.mysite.com/view.php?id={$query}"> OPT; It does not display the number in the link. What have i done wrong? Link to comment https://forums.phpfreaks.com/topic/253132-select-random-id-from-table-to-link/ Share on other sites More sharing options...
marcelobm Posted December 14, 2011 Share Posted December 14, 2011 try this $query_result = mysql_result($query,0); echo <<<OPT <meta http-equiv="REFRESH" content="0;url=http://www.mysite.com/view.php?id={$query_result}"> OPT; Link to comment https://forums.phpfreaks.com/topic/253132-select-random-id-from-table-to-link/#findComment-1297815 Share on other sites More sharing options...
starvinmarvin14 Posted December 14, 2011 Author Share Posted December 14, 2011 $query=mysql_query("SELECT id FROM image WHERE active='yes' ORDER BY RAND() LIMIT 1"); $query_result = mysql_result($query,0); echo <<<OPT <meta http-equiv="REFRESH" content="0;url=http://www.mysite.com/view.php?id={$query_result}"> OPT; Tried that and got this... "Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/.../public_html/.../random.php on line 9" Link to comment https://forums.phpfreaks.com/topic/253132-select-random-id-from-table-to-link/#findComment-1297890 Share on other sites More sharing options...
marcelobm Posted December 14, 2011 Share Posted December 14, 2011 it seems like your query is not returning any rows, try to do this if(mysql_num_rows($query)){ $query_result = mysql_result($query,0); echo <<<OPT <meta http-equiv="REFRESH" content="0;url=http://www.mysite.com/view.php?id={$query_result}"> OPT; }else{ echo 'no records found'; } Link to comment https://forums.phpfreaks.com/topic/253132-select-random-id-from-table-to-link/#findComment-1297894 Share on other sites More sharing options...
Pikachu2000 Posted December 14, 2011 Share Posted December 14, 2011 No, the query is completely failing. Use mysqli_error to get some insight as to why that's happening. Link to comment https://forums.phpfreaks.com/topic/253132-select-random-id-from-table-to-link/#findComment-1297896 Share on other sites More sharing options...
starvinmarvin14 Posted December 14, 2011 Author Share Posted December 14, 2011 Tried this... $query = mysql_query("SELECT id FROM image WHERE active='yes' ORDER BY RAND() LIMIT 1"); if(mysql_num_rows($query)){ $query_result = mysql_result($query,0); echo <<<OPT <meta http-equiv="REFRESH" content="0;url=http://www.mysite.com/view.php?id={$query_result}"> OPT; }else{ echo 'no records found'; } Now I am getting this error... Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/.../public_html/.../random.php on line 9 no records found I only have to rows at the moment with the id 18 and 19. How would i use mysqli_error()? Link to comment https://forums.phpfreaks.com/topic/253132-select-random-id-from-table-to-link/#findComment-1297899 Share on other sites More sharing options...
mikosiko Posted December 14, 2011 Share Posted December 14, 2011 for your code sure Pikachu2000 means to say mysql_error() instead of mysqli_error() Link to comment https://forums.phpfreaks.com/topic/253132-select-random-id-from-table-to-link/#findComment-1297900 Share on other sites More sharing options...
Pikachu2000 Posted December 14, 2011 Share Posted December 14, 2011 Yes, you're right . . . I thought I saw an i in there, but I guess not. Time for new glasses, I suppose. Link to comment https://forums.phpfreaks.com/topic/253132-select-random-id-from-table-to-link/#findComment-1297904 Share on other sites More sharing options...
starvinmarvin14 Posted December 14, 2011 Author Share Posted December 14, 2011 $row = mysql_query("SELECT * FROM image ORDER BY RAND() LIMIT 1;"); $id = $row['id']; echo <<<OPT <meta http-equiv="REFRESH" content="0;url=http://www.mysite.com/view.php?id={$id}"> OPT; echo $row When i echo the row I get "Resource id #2" every time. Link to comment https://forums.phpfreaks.com/topic/253132-select-random-id-from-table-to-link/#findComment-1297925 Share on other sites More sharing options...
Pikachu2000 Posted December 14, 2011 Share Posted December 14, 2011 $row holds the result resource returned by mysql_query. You'll need to use one of the mysql_fetch_* functions, such as mysql_fetch_assoc to extract the data from the result resource. Link to comment https://forums.phpfreaks.com/topic/253132-select-random-id-from-table-to-link/#findComment-1297927 Share on other sites More sharing options...
starvinmarvin14 Posted December 14, 2011 Author Share Posted December 14, 2011 Thanks this worked! $row = mysql_fetch_array(mysql_query("SELECT id FROM image ORDER BY RAND() LIMIT 1")); $id = $row['id']; echo <<<OPT <meta http-equiv="REFRESH" content="0;url=http://www.mysite.com/view.php?id={$id}"> OPT; Link to comment https://forums.phpfreaks.com/topic/253132-select-random-id-from-table-to-link/#findComment-1297937 Share on other sites More sharing options...
Pikachu2000 Posted December 14, 2011 Share Posted December 14, 2011 It may work, but it's a really bad way to do it with all the functions nested like that. It makes debugging that part of the code quite difficult. Link to comment https://forums.phpfreaks.com/topic/253132-select-random-id-from-table-to-link/#findComment-1297940 Share on other sites More sharing options...
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