Glenskie Posted April 3, 2014 Share Posted April 3, 2014 I am running this query and the only thing i want is the title and for some reason it is giving me this. Recourse id # 7 ? Here is my query that i am running. $photo = $_GET['image']; $retrievetitle = "SELECT title FROM photo_from_user WHERE picture = '$photo' limit 1"; $title = mysql_query($retrievetitle); Link to comment https://forums.phpfreaks.com/topic/287495-query-returning-resource-id-7/ Share on other sites More sharing options...
gristoi Posted April 3, 2014 Share Posted April 3, 2014 mysql_query returns a resource, you need to loop through that resource to get the data back. Link to comment https://forums.phpfreaks.com/topic/287495-query-returning-resource-id-7/#findComment-1474829 Share on other sites More sharing options...
ginerjm Posted April 3, 2014 Share Posted April 3, 2014 To finish your (deprecated) code: if ($title) { $row = MySQL_fetch_assoc($title); $my_title = $row['title']; } else { echo "Error running title query - message is " . MySQL_error(); exit(); } Of course you should change your db interface to mysqlI or pdo to keep up with the times. Link to comment https://forums.phpfreaks.com/topic/287495-query-returning-resource-id-7/#findComment-1474830 Share on other sites More sharing options...
Psycho Posted April 3, 2014 Share Posted April 3, 2014 Or more simply use mysql_result(). I've renamed the variables to more commonly used for the intended purpose. There's no need to give all the query strings or the results unique names: $photo = mysql_real_escape_string(trim($_GET['image'])); $query = "SELECT title FROM photo_from_user WHERE picture = '$photo' limit 1"; $result = mysql_query($query); if(!$result) { echo "Error running title query - message is " . mysql_error(); exit(); } $title = mysql_result($result, 0); Link to comment https://forums.phpfreaks.com/topic/287495-query-returning-resource-id-7/#findComment-1474833 Share on other sites More sharing options...
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