Sethm Posted January 6, 2010 Share Posted January 6, 2010 I have some code that is using a second database to store the image files for the particular listing in my database. What I am wondering is how to find out how to use an if statement on if there is anything pulled or not when I pull the information from it using the following code: $sql = mysql_query("SELECT * FROM imagedb WHERE listingsid=$id"); I've tried the statement: if ($sql != "") { echo "information"; } It didn't work... Could someone help point me in the right direction? Thanks a lot! Link to comment https://forums.phpfreaks.com/topic/187485-display-based-on-secondary-database-information-being-present-or-not/ Share on other sites More sharing options...
premiso Posted January 6, 2010 Share Posted January 6, 2010 I would suggest using mysql_num_rows if (mysql_num_rows($sql) > 0) { //echo "information returned."; } Should work for you. Link to comment https://forums.phpfreaks.com/topic/187485-display-based-on-secondary-database-information-being-present-or-not/#findComment-989956 Share on other sites More sharing options...
Sethm Posted January 6, 2010 Author Share Posted January 6, 2010 Thank you SOO much for your quick response - saved my hair! Link to comment https://forums.phpfreaks.com/topic/187485-display-based-on-secondary-database-information-being-present-or-not/#findComment-989958 Share on other sites More sharing options...
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