darksniperx Posted November 26, 2007 Share Posted November 26, 2007 $result = mysql_query("SELECT * FROM feed WHERE feed_id = '$id'"); if(!$result) { die("query unsuccessfull"); } else { echo'queury successfull'; } something along these lines, but to be able to check any kind of query, add, delete, update,... the example above does not work me. Link to comment https://forums.phpfreaks.com/topic/78862-solved-checking-if-sql-queury-was-successfull/ Share on other sites More sharing options...
revraz Posted November 26, 2007 Share Posted November 26, 2007 What do you want to know, if rows were found or if a connection was made to the DB? Link to comment https://forums.phpfreaks.com/topic/78862-solved-checking-if-sql-queury-was-successfull/#findComment-399144 Share on other sites More sharing options...
darksniperx Posted November 26, 2007 Author Share Posted November 26, 2007 if query was successfull, meaning there were no errors during query, and if rows were found or not Link to comment https://forums.phpfreaks.com/topic/78862-solved-checking-if-sql-queury-was-successfull/#findComment-399222 Share on other sites More sharing options...
trq Posted November 26, 2007 Share Posted November 26, 2007 <?php $sql = "SELECT * FROM feed WHERE feed_id = '$id'"; if ($result = mysql_query($sql)) { // query was successfull. if (mysql_num_rows($result)) { // rows were found, fetch array and display result. } else { echo "No rows found"; } } else { echo "Query failed<br />$sql<br />" . mysql_error(); } ?> Link to comment https://forums.phpfreaks.com/topic/78862-solved-checking-if-sql-queury-was-successfull/#findComment-399287 Share on other sites More sharing options...
darksniperx Posted November 26, 2007 Author Share Posted November 26, 2007 thx, that is all that I needed! Link to comment https://forums.phpfreaks.com/topic/78862-solved-checking-if-sql-queury-was-successfull/#findComment-399314 Share on other sites More sharing options...
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