try this: (I don't have experience w/ mysql, but I do with other dbs)

<select name="test drop down">
$sql_select = "select * from dropdown_options";
$results = mysql_query($sql_select);
while ($row = mysql_fetch_array($results);
  echo "<option VALUE=".$row["col1"].">".$row["col2"];

-----Original Message-----
From: Matthew Darcy [mailto:[EMAIL PROTECTED]]
Sent: Monday, February 25, 2002 10:06 AM
Subject: [PHP] RE: (HTML related) helping to creating dropdown lists
from fetched arrays.

it does help a little.

I know the html is wrong I used this as an example to what I wanted.

What I need to know is how to use PHP to generate the options in the list
from col1 and show the option 2.
The data for these is got from an array.

so really what I want to know is how to code in PHP

<select name=test drop down>
  <option value=col1($row1)>col2($row1)
  <option value=col1($row2)>col2($row2)
  <option value=col1($row3)>col2($row3)

you get the idea ?



-----Original Message-----
From: Cece [mailto:[EMAIL PROTECTED]]On Behalf Of Heilig (Cece) Szabolcs
Sent: 24 February 2002 23:07
Subject: OT:(HTML related) helping to creating dropdown lists from
fetched arrays.

> I want to create a dropdown list with options from a table.

> <?php
> $sql_select = "select * from dropdown_options";
> $results = mysql_query($sql
> _select);
> while ($row = mysql_fetch_array($results);
> {
>       echo "<option VALUE=$row["col1"] NAME=option1>"
> }


The problem is simple: wrong HTML
You have to make html code similar that:

<select name=gender>
  <option value=F>female
  <option value=M>male
  <option value=0>not known :)

Hope it helps


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