Populating a Dropdown

[Anon]

What is the easiest way to populate a drop down with values from a database? I have tried all types of code but they are not working for me.

Thanks in advance.

Jason Hall

[Anon]

make a function like this:

function populateDropdown($value, $name) {
return "<option value=\"".$value."\">".$name."</option>";
}

Then in your code do where you have your sql result set:

$html = "<select name=\"test\">";
while($row = mysql_fetch_array($result)
$html .= populateDropdown($row['value'], row['name']);
$html .= "</select>";

[Anon]

I have table called armor, so u have to change that.

<SELECT NAME="dropdown">
<?
$choose=mysql_query("SELECT name FROM armor");

while($armorname = mysql_fetch_array($choose))
{
print("<OPTION VALUE=\"$armorname[2]\">$armorname[2]</OPTION>\n");
}
?>
</SELECT>

this should work, just change some names and fields and done. U use array[x] to indicate which field u want

hopt this help

TidalWave

[kamy99]

How would you make a specific code "selected"
in this code?

[Anon]

I did that and I get this error

Parse error: parse error in /home/webfame/hraashf-www/athletes.php on line 50

Here is my code
<?php
include("includes/connect.php");

$query = "SELECT Distinct YearInducted from athletes order by YearInducted ASC";

$result = mysql_query($query,$link);
function populateDropdown($value, $name) {
return "<option value=\"".$value."\">".$name."</option>";
}

html = "<select name=\"Year\">";
while($row = mysql_fetch_array($result)
$html .= populateDropdown($row["YearInducted"], $row["YearInducted"]);
$html .= "</select>";
?>

[Anon]

not, but he didnt ask for that

[kamy99]

I know, but I've been trying to get this thing done for a while now without much success! Perhaps you can help?
I'm using the code below to populate a selction box with States. It works fine.
What I need is to get the appropriate item selected.
On the page prior to this, users have already logged in and there's a State column for the users. So all I need is to get the user's state pre-selected. ie if user is from California, California gets selected.
The snippet is pasted below.

Thanx in advance........
Kamy

{
include_once "common.inc";
$CountryCode = "US";
$result = mysql_query("select * from provinces where CY='$CountryCode'");
while ($query_data = mysql_fetch_row($result))
{
echo "<option Value=",$query_data[2]," >",$query_data[1],"</option>";
}

[Anon]

Tidalwave,

Thanks a million. Works like a charm. I really really appreciate it.

Thanks
Jason Hall

[Anon]

there is one ')' missing here:

while($row = mysql_fetch_array($result)

should be

while($row = mysql_fetch_array($result))

[Anon]

Can u clearify a bit more, i'm not from english so it's my second language and I dont quite get ur question..

I'll do my best to help ya

[kamy99]

The selection shows State names. For example; California, New Jersey, Florida....

What I need is this:
If user is from California, make California selected. If user is from Florida, make Florida selected....etc

I hope this is clear enough.

[Anon]

Its a hard one, and I dont have a script lying around for it, but if u ask me

use php to write the forms, then when u come to the point of <option quit here and write an if statement to select what to write next, if the value of the option = the "user.state" then write selected there too if not just write the value.

I hope this help, cant help quite more now since i never really tried it or have more coded available

this is what I would try, Hope it help

TidalWave

[kamy99]

Actually I just solved it! I ran a second query, where the state was equal to user.state (just like you mentioned).

Thanx......

[Anon]

glad to hear u found it, but dont thank me, i didnt do anything