mastubbs Posted July 7, 2013 Share Posted July 7, 2013 Hi all, Im trying to populate a drop down list from a php table but for some reason im getting stuck. The table structure is: Table name: Par1AddPatients_dynlist_items Columns: ID, listid, name, value I am trying to populate the list with names and corresponding values from the table, where the list id is ‘1’. <select name="ward_list"> <?php $connect = mysql_connect("localhost","username ","password "); if (!$connect) { die("MySQL could not connect!"); } $DB = mysql_select_db('jasperss_par1pats'); if(!$DB) { die("MySQL could not select Database!"); } $ward_list = mysql_query("SELECT name AND value FROM Par1AddPatients_dynlist_items WHERE listid = '1'"); while ($row = mysql_fetch_array($ward_list)){ echo '<option value="'. $row['value'] .'">'. $row['name'] .'</option>'; } ?> </select> At the moment I’m just getting an empty drop-down box. Any ideas where i'm going wrong? Thanks in advance for any help with this, Matt Quote Link to comment Share on other sites More sharing options...
mac_gyver Posted July 7, 2013 Share Posted July 7, 2013 in programming, AND is a logical operator. name AND value produces the boolean result of the column value in name AND'ed with the column value in value. to make a list of columns to SELECT, you would separate them with commas , Quote Link to comment Share on other sites More sharing options...
thara Posted July 7, 2013 Share Posted July 7, 2013 As @mac_gyver has said, your select query is wrong. Here is generic SQL syntax of SELECT command to fetch data from MySQL table: SELECT field1, field2,...fieldN FROM table_name [WHERE Clause] Read it from manual Quote Link to comment Share on other sites More sharing options...
Solution thara Posted July 7, 2013 Solution Share Posted July 7, 2013 Try this. <select name="ward_list"> <?php $connect = mysql_connect("localhost","username ","password "); if (!$connect) { die("MySQL could not connect!"); } $DB = mysql_select_db('jasperss_par1pats'); if(!$DB) { die("MySQL could not select Database!"); } $ward_list = mysql_query("SELECT name, value FROM Par1AddPatients_dynlist_items WHERE listid = '1'"); while ($row = mysql_fetch_array($ward_list)){ echo '<option value="'. $row['value'] .'">'. $row['name'] .'</option>'; } ?> </select> Side note: Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated, read this . Learn about prepared statements instead, and use PDO or MySQLi Quote Link to comment Share on other sites More sharing options...
mastubbs Posted July 7, 2013 Author Share Posted July 7, 2013 Ah ok i see what you mean. Will try to get into using prepared statements. Thanks for the help. Matt Quote Link to comment Share on other sites More sharing options...
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