problem populating dropdown menu

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September 19th, 2004, 04:30 PM

Gerald

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Join Date: Sep 2004

Posts: 2

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problem populating dropdown menu

Hi!!

I am trying to populate my dropdown menu by getting the values
from the database.In this code I am trying to populate the "name" field
from the table "details".I am getting a blank dropdown without any values in it.
Can anybody please help me in this code.

<html>
<body>
<?php
include ("localhostconnection.php");
?>
<select name="id" style="width: 200px;">
<?php
$query = mysql_query("SELECT name FROM details" );
while ($row = mysql_fetch_array($query))
{
$j = mysql_num_fields($query);
for($i=0;$i<$j;$i++)
{
$k = mysql_field_name($query,$i);
$$k = $row[$k];
}
print("<option value=\"$id\">$user</option>\n");
}
?>
</select>
</body>
</html>

Thankyou.

September 19th, 2004, 06:20 PM

kode_monkey

Contributing User

Join Date: Jul 2003

Posts: 367

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OK based on what I see here I would highly recommend looking through a few php tutorials. There's loads out there and they will help you understand what parts of this script do.

To give you some help to get you started though the for loop in your code is of no use at all. You just need the while loop with the print statement in it. Something like

// run query

// while (get a result)

// print values

-KM-

September 19th, 2004, 07:10 PM

Gerald

Registered User

Join Date: Sep 2004

Posts: 2

Time spent in forums: < 1 sec
Reputation Power: 0

Thankyou

Hi !!

I think i was wrong in mentioning the field names and I think we need a pointer for sure
which points to the each item in dropdown list.I didn't understand that.Now i got it.
Anyway Thankyou so.... much

September 20th, 2004, 10:59 AM

MadCowDzz

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Dev Articles Frequenter (2500 - 2999 posts)

Join Date: Jan 2003

Location: Toronto, Canada

Posts: 2,890

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Where's $id being set/changed?

You could change all that PHP to read:

PHP Code:


		
			
 $query = mysql_query("SELECT name FROM details" );

while ($row = mysql_fetch_array($query))

{

     $id++;

     print("<option value=\"$id\">".$row['name']."</option>\n");

}

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