Drop Down list from database in PHP/ NOT WORKING ........HELP

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prajita

Oct 27th 2007, 4:34 pm

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

nico_swd

Oct 27th 2007, 4:36 pm

Your question has been answered here?

http://forums.digitalpoint.com/showthread.php?t=530644

:confused:

prajita

Oct 27th 2007, 5:03 pm

Nope its not working.....

Dirty-Rockstar

Oct 27th 2007, 5:12 pm

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. :D works wonders for me

nico_swd

Oct 27th 2007, 5:14 pm

Nope its not working.....

I replied to your other thread. Did you try what I said?

And please be more specific. What exactly "does not work"?

prajita

Oct 27th 2007, 5:17 pm

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">

</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

nico_swd

Oct 27th 2007, 5:26 pm

I cannot populate the stu_id from the database into my drop down list

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());

?>

prajita

Oct 27th 2007, 5:32 pm

I really appreciate....

The code works

What did you do?

You are a magician....

Thanks

nico_swd

Oct 27th 2007, 5:36 pm

Actually, there's a slight typo:

print '<select name=\"studentid"\>';

Should be:

print '<select name="studentid">';

(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.

prajita

Oct 27th 2007, 6:23 pm

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">

</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>

prajita

Oct 27th 2007, 6:26 pm

thank you anyways..........

I figured out the mistake i made.

Thanks