tobeyt23 Posted August 3, 2007 Share Posted August 3, 2007 I am calling all fields from a table, just to many to call the ones i need. I would like to drop the first 2 from being shown, how can I do that by the code below or do I need to rewrite completely? while ($line = mysql_fetch_array($store_data, MYSQL_ASSOC)) { foreach ($line as $data) { echo $data . ', '; } } Thanks Quote Link to comment https://forums.phpfreaks.com/topic/63195-solved-database-display-help/ Share on other sites More sharing options...
tobeyt23 Posted August 3, 2007 Author Share Posted August 3, 2007 I got it while ($line = mysql_fetch_array($store_data, MYSQL_ASSOC)) { foreach ($line as $key => $value) { if($key != 'id' && $key != 'applicant_id_FK') { echo $value . ', '; } } } Quote Link to comment https://forums.phpfreaks.com/topic/63195-solved-database-display-help/#findComment-314974 Share on other sites More sharing options...
sKunKbad Posted August 3, 2007 Share Posted August 3, 2007 <?php $result = mysql_query("SELECT id, name, salary FROM employees", $conn); while (list($id, $name, $salary) = mysql_fetch_row($result)) { echo " <tr>\n" . " <td><a href=\"info.php?id=$id\">$name</a></td>\n" . " <td>$salary</td>\n" . " </tr>\n"; } ?> just skip a variable in the list Quote Link to comment https://forums.phpfreaks.com/topic/63195-solved-database-display-help/#findComment-314975 Share on other sites More sharing options...
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