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Thread: Drop down box
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May 19, 2005, 06:22 #1
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Drop down box
I want to have a drop down box where a user can choose an address. I got 10 different addresses. Problem with my code now is that its displaying 10 different drop down down boxes instead of displaying 1 with all 10 addresses in. Somebody who can help?
PHP Code:echo'
<form method="post">
Sorter etter:<br>
<select name="sort" size="1">
<option value="test"> '.$v_address.' </option>
</select>
<input type="submit" name="Request" value="OK">
</form>';
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May 19, 2005, 06:27 #2
if the addresses are in an array you can foreach over the array and output the "<option>" parts of the form
PHP Code:<form method="post">
Sorter etter:<br>
<select name="sort" size="1">
<?php
foreach($addressArray as $v_address){
print '<option value="test"> '.$v_address.' </option>'
}
?>
</select>
<input type="submit" name="Request" value="OK">
</form>Erh
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May 19, 2005, 06:29 #3
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Its fetched as object. Can i do something similar then?
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May 19, 2005, 06:33 #4
Originally Posted by sirikit
Erh
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May 19, 2005, 06:38 #5
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Hehe, im having dinner soon
I have a database with all the addresses stored. So im fetching my database and table as object like this:
PHP Code:$q = "SELECT *, leilighet.leilighetsNr AS lNr FROM leilighet LEFT JOIN leietager ON leietager.leilighetsNr = leilighet.leilighetsNr
WHERE leietager.leilighetsNr IS NULL ORDER BY $order LIMIT $from, $max_results";
$result = mysql_query($q, $conn) or die(mysql_error());
while ($row = mysql_fetch_object($result))
{
$row_color = ($row_count % 2) ? $color1 : $color2;
// Henter data fra tabeller i databasen. De blir da lagret til egne variabler.
$v_photo=$row->bilde_url;
$v_overskrift=$row->overskrift;
$v_adresse=$row->leiladresse;
$v_post=$row->leilpostnr;
$v_sted=$row->leilpoststed;
$v_kvm=$row->kvm;
$v_rom=$row->antallRom;
$v_pris=$row->pris;
$v_number=$row->lNr;
?>
<form method="post">
Sorter etter:<br>
<select name="sort" size="1">
<?php
foreach($addressArray as $v_adresse)
{
print '<option value="test"> '.$v_adresse.' </option>';
}
?>
</select>
<input type="submit" name="Request" value="OK">
</form>
<?
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May 19, 2005, 06:51 #6PHP Code:
<form method="post">
Sorter etter:<br>
<select name="sort" size="1">
<?php
$q = "SELECT *, leilighet.leilighetsNr AS lNr FROM leilighet LEFT JOIN leietager ON leietager.leilighetsNr = leilighet.leilighetsNr
WHERE leietager.leilighetsNr IS NULL ORDER BY $order LIMIT $from, $max_results";
$result = mysql_query($q, $conn) or die(mysql_error());
while ($row = mysql_fetch_object($result))
{
$row_color = ($row_count % 2) ? $color1 : $color2;
// Henter data fra tabeller i databasen. De blir da lagret til egne variabler.
$v_photo=$row->bilde_url;
$v_overskrift=$row->overskrift;
$v_adresse=$row->leiladresse;
$v_post=$row->leilpostnr;
$v_sted=$row->leilpoststed;
$v_kvm=$row->kvm;
$v_rom=$row->antallRom;
$v_pris=$row->pris;
$v_number=$row->lNr;
print '<option value="test"> '.$v_adresse.' </option>';
}
?>
</select>
<input type="submit" name="Request" value="OK">
</form>Oooops some is still in norwegian...
It's perfectly understandable....
Okay so you just use the loop you're already using with the DB result set. But you put the form tags and the select tags outside the loop so your just outputing the option tags in the loop. HTH (and I hope the code works...sorry if it's not quite right...it's the idea that counts.)
Erh
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May 19, 2005, 08:07 #7
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NICE!! works perfect. thanks
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