Please help me with the code below: echo "<select name=\"studentid\">"; while($row = mysql_fetch_row($results_id)); { echo "<option value=\"".$row['stu_id']."\">".$row['stu_id']."</option>"; } echo "</select>"; print '</select>'; In above code i am trying to create a drop down list from a data in a table in a database The name of the field in table is stu_id Please help I really need your help
debug. echo or print out the vars, and then write if statements if (isset($row)){ print "worked"; }else{ print "KABLAMMO"; } checking to see if all vars and queries are working is the first way to start. works wonders for me
I replied to your other thread. Did you try what I said? And please be more specific. What exactly "does not work"?
I cannot populate the stu_id from the database into my drop down list Okay here is the total code: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title> Search Results </title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <style type="text/css"> <!-- span.c1 {color: blue; font-size: 120%} span.c2 { color: red; font-style:italic } --> </style> </head> <body> <?php $host= "localhost"; $user = "---------"; $passwd = "---------"; $database = "------------"; $connect = @mysql_connect($host, $user, $passwd) or die("connect error: " . mysql_error()); $table_name = 'student'; @mysql_select_db($database) or die('select_db error: ' . mysql_error()); print "<span class=\"c1\">$table_name Data</span><br>"; $query = "SELECT stu_id FROM $table_name"; print "The query is <span class=\"c2\"> $query </span> <br>"; $results_id=@mysql_query($query, $connect) or die('query error: ' . mysql_error()); print '<select name=\"studentid"\>'; while ($row = mysql_fetch_assoc($result,MYSQL_ASSOC)) { print '<option name=\"stu_id\"> $row["stu_id"]</option>'; } print '</select>'; @mysql_close($connect) or die('close error: ' . mysql_error()); ?> </body> </html> I tried so many times but i still cannot display values in my drop down list. i would appreciate your help
I figured as much. What I meant was, what does actually happen? Any PHP errors (if so, post them), are the options just in blank? Anyway, try this. I couldn't test it but I don't see why it wouldn't work. <?php $host= "localhost"; $user = "---------"; $passwd = "---------"; $database = "------------"; $table_name = 'student'; $connect = @mysql_connect($host, $user, $passwd) or die("connect error: " . mysql_error()); @mysql_select_db($database) or die('select_db error: ' . mysql_error()); print "<span class=\"c1\">$table_name Data</span><br>"; $query = "SELECT stu_id FROM $table_name"; $results = @mysql_query($query, $connect) or die('query error: ' . mysql_error()); print '<select name=\"studentid"\>'; while ($row = mysql_fetch_assoc($result)) { print "<option name=\"{$row['stu_id']}\">{$row['stu_id']}</option>\n"; } print '</select>'; @mysql_close($connect) or die('close error: ' . mysql_error()); ?> PHP:
Actually, there's a slight typo: print '<select name=\"studentid"\>'; PHP: Should be: print '<select name="studentid">'; PHP: (No need to escape the double quotes, in a single quoted string.) And compare the 2 codes carefully to see the changes I made. Oh, and one last suggestion: Don't be afraid of wasting lines. They're for free. Plus the code will be MUCH more readable and it's easier to spot errors this way.
I am sorry to bug you like this but i really need help. this is my school project and i am trying my best. I already finished more than 70% and i am stuck on this. I need to do the similar thing (creating drop down list from values in a table in a database) in the same page from another table named "course" and the name of attribute is "course_num". I copied the same exact code from above(which is working and changed the name of all the required variable but the new drop down is not working........ <head> <style type="text/css"> <!-- span.c1 {color: blue; font-size: 120%} span.c3{ font-size:175%} span.c4(text-align=center; font-size:175%} span.c2 { color: red; font-style:italic } --> </style> </head> <body> <?php $host= "localhost"; $user = "rkarki"; $passwd = "Muna1!"; $database = "rkarki"; $connect = @mysql_connect($host, $user, $passwd) or die("connect error: " . mysql_error()); $table_name = 'student'; @mysql_select_db($database) or die('select_db error: ' . mysql_error()); print "<span class=\"c1\">$table_name Data</span><br>"; $query = "SELECT stu_id FROM $table_name"; print "The query is <span class=\"c2\"> $query </span> <br>"; $results_id=@mysql_query($query, $connect) or die('query error: ' . mysql_error()); print "<span class=\"c3\"> Student ID</span> <br>"; print '<select name="studentid">'; while ($row = mysql_fetch_assoc($results_id)) { print "<option name=\"{$row['stu_id']}\">{$row['stu_id']}</option>\n"; } print '</select>'; @mysql_close($connect) or die('close error: ' . mysql_error()); print "<span class=\"c4\"> Course Number</span> <br>"; $query = "SELECT cours_num FROM course"; $results_id=@mysql_query($query, $connect) or die('query error: ' . mysql_error()); print '<select name="CourseID">'; while ($row = mysql_fetch_assoc($results_id1)) { print "<option name=\"{$row['cours_num']}\">{$row['cours_num']}</option>\n"; } print '</select>'; print '</span>'; @mysql_close($connect) or die('close error: ' . mysql_error()); ?> </body> </html>