jordanwhite Posted December 19, 2010 Share Posted December 19, 2010 Hey guys, im new around here, and i have just recently started teaching myself PHP, by looks of this site i seem very experienced but im eager to learn I seem to get an error with this code. The point of this code is to look into my database and pull out all entries with the first name _____ (whatever the user inputs) Error: Warning: mysql_fetch_array() expects parameter 1 to be resource, null given in C:\xampp\htdocs\searchans.php on line 12 Code: Filename:search.php <html> <form action='searchans.php' method='POST'> First Name: <input type='text' name='firstname'><br> <input type='submit' value='Search'> </form> </html> Code: Filename:searchans.php <?php $firstname = $_POST['firstname']; $con = mysql_connect("localhost","peter","abc123"); if (!$con) { die('Could not connect: ' . mysql_error()); } $query = mysql_query("SELECT * FROM Persons WHERE FirstName='$firstname'"); while($row = mysql_fetch_array($result)) { echo $row['FirstName'] . " " . $row['LastName'] . " " . $row['Age']; echo "<br />"; } ?> Link to comment https://forums.phpfreaks.com/topic/222116-basic-searching-database-for-entry/ Share on other sites More sharing options...
johnny86 Posted December 19, 2010 Share Posted December 19, 2010 while($row = mysql_fetch_array($result)) SHOULD BE: while($row = mysql_fetch_array($query)) Link to comment https://forums.phpfreaks.com/topic/222116-basic-searching-database-for-entry/#findComment-1149171 Share on other sites More sharing options...
jordanwhite Posted December 19, 2010 Author Share Posted December 19, 2010 changed the code still getting error Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\searchans.php on line 15 Link to comment https://forums.phpfreaks.com/topic/222116-basic-searching-database-for-entry/#findComment-1149172 Share on other sites More sharing options...
johnny86 Posted December 19, 2010 Share Posted December 19, 2010 Try adding this: $query = mysql_query("SELECT * FROM Persons WHERE FirstName='$firstname'"); //Add this below if(!$query) die("MySQL query failed!"); Link to comment https://forums.phpfreaks.com/topic/222116-basic-searching-database-for-entry/#findComment-1149173 Share on other sites More sharing options...
jordanwhite Posted December 19, 2010 Author Share Posted December 19, 2010 the error just came up instantly "MySQL query failed!" Link to comment https://forums.phpfreaks.com/topic/222116-basic-searching-database-for-entry/#findComment-1149174 Share on other sites More sharing options...
johnny86 Posted December 19, 2010 Share Posted December 19, 2010 Then you have to check your mysql query. Obviously you have a connection. Oh and stupid me.. You have to select a database... <?php $firstname = $_POST['firstname']; $con = mysql_connect("localhost","peter","abc123"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("your_database_name"); $query = mysql_query("SELECT * FROM Persons WHERE FirstName='$firstname'"); while($row = mysql_fetch_array($query)) { echo $row['FirstName'] . " " . $row['LastName'] . " " . $row['Age']; echo "<br />"; } ?> Link to comment https://forums.phpfreaks.com/topic/222116-basic-searching-database-for-entry/#findComment-1149175 Share on other sites More sharing options...
jordanwhite Posted December 19, 2010 Author Share Posted December 19, 2010 wow, cant believe I forgot that. Thanks for the help though buddah Link to comment https://forums.phpfreaks.com/topic/222116-basic-searching-database-for-entry/#findComment-1149176 Share on other sites More sharing options...
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