I assume you want the array populated with something other than the
display name in the <select>, so I've made up "OrgId" as a field name in
your database. How 'bout something a little easier to read, using a
more consistent style?
<?php
if (!$db=mysql_connect("localhost","lodestone","trypass"))
$err=mysql_error();
if (!$err) if (!mysql_select_db("catapult_com"))
$err=mysql_error();
if (!$err) {
$q="SELECT OrgId,OrgName FROM TableResults ORDER BY OrgName";
if ($r=mysql_query($q,$db)) {
$options=array(); // for clarity
while ($row=mysql_fetch_array($r))
$options[$row["OrgId"]]=$row["OrgName"];
} else
$err=mysql_error();
}
// If all's well so far, show the <select>...
if ($err)
print "Database error: <tt><b>" . $err . "</b></tt>";
else {
print "<select name='OrgName[]' size=5 id='OrgName'>\n"
foreach ($options as $id => $name)
printf("\t<option value='%s'>%s\n",$id,$name);
print "</select>\n";
}
?>
Note that none of this code has been tested. :)
p
On Sun, Dec 29, 2002 at 01:18:22AM +0800, Denis L. Menezes wrote:
>
> <select name="OrgName[]" size=5 id="OrgName" >
> <?php
> //connecting to the database
> $link = mysql_connect("localhost","lodestone","trypass");
> if ($link){
> Print "";
> } else {
> Print "No connection to the database";
> }
> if (!mysql_select_db("catapult_com")){
> Print "Couldn't connect database";
> } else {
> Print ""."<br>\n";
> }
>
> $sql="SELECT OrgName From TableResults ORDER BY OrgName";
> $result=mysql_query($sql);
>
> While($Organisation=mysql_fetch_array($result))
> {
> Print("<OPTION VALUE=\"$Organisation[0]\">$Organisation[1]\n");
> }
>
> ?>
> </select>
>
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--
Paul Chvostek <[EMAIL PROTECTED]>
Operations / Abuse / Whatever
it.canada, hosting and development http://www.it.ca/
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