PHP dropdown with items from database

Question

I can't seem to get values of a table in my dropdownlist. I get no errors, only an empty dropdownlist. This is my code:

<?php               
include_once ("classes/Keten.class.php");
$keten = new Keten();
$allKet = $keten->getAllKetens();

echo '<select name="fk_keten_id">';             
while($row = mysql_fetch_array($allKet))
{
  echo '<option value="' . 
       $row['pk_keten_id'] . '">' . $row['keten_naam'] . '</option>';
}
echo '</select>'; 
?>

Function getAllKetens works, because I use it in an other page.

Thanks in advance :)

on request, the getAllKet() function in classes/Keten.class.php

public function getAllKetens()
    {
        include ("connection.php");
        $sSql = "SELECT * FROM tblKetens;";
        $vResult = mysqli_query($link, $sSql);
        mysqli_close($link);
        return($vResult);
    }

Could you debug $allKet and let us know what you get?

RabidFire
– RabidFire

2010-12-01 18:54:35 +00:00
Commented Dec 1, 2010 at 18:54 — RabidFire
– RabidFire, Commented Dec 1, 2010 at 18:54
do print_r($allket); what does it say?

Breezer
– Breezer

2010-12-01 18:56:53 +00:00
Commented Dec 1, 2010 at 18:56 — Breezer
– Breezer, Commented Dec 1, 2010 at 18:56

brett · Accepted Answer · 2010-12-01 19:35:48Z

2

I'm not sure mixing mysqli and mysql commands can be done, in your fetch_array you used mysql_fetch_array instead of mysqli_fetch_array.

answered Dec 1, 2010 at 19:35

brett

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Comments

snarkback · Accepted Answer · 2010-12-01 19:41:36Z

1

Your function getAllKetens() is closing the database. You shouldn't be trying to fetch a row after closing the connection to the database.

Soap box aside: It might be nice to have some more standard naming convention going on in there, as well. If you're returning a handle to a mysql query object, consider $query or $hQuery instead of what's (to me) a lot more cryptic as $allKet. IMO, if a function is called "getAllKetens()" it should return an array of data, not a handle to a query.

answered Dec 1, 2010 at 19:41

snarkback

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1 Comment

snarkback Over a year ago

addendum: though you should definitely close the connection, but do it after you're done with all query resources used against that connection (your $link)

Byron Whitlock · Accepted Answer · 2010-12-01 18:58:57Z

0

Make sure errors are turned on
View the source, If there are any errors, they will be hidden by the select tag
Do a print_r($allket); to see if getAllKeten() actually returned anything.

answered Dec 1, 2010 at 18:58

Byron Whitlock

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2 Comments

KawaGreen Over a year ago

with the print_r($allket); I get a "1"

casablanca Over a year ago

@Glenn: Then you need to check your getAllKetens function -- maybe post the code in your question.

Ian · Accepted Answer · 2010-12-01 19:11:52Z

0

I suspect the $allKet is not what you think it is.

First, check the HTML. If you are getting <select name="???"></select> then

a) Check $allKet is a resouce (I think print_r will tell you). b) Check that the SQL is selecting rows. (try it in phpmyadmin) c) Check that you have correct spellings for field names.

If you are seeing <option value="??"></option> entries, then

a) Check the field names you are printing are correct b) Check you are not hiding them with an alias (the print will show the last field).

Turn error reporting up (in php.inf) and remove any errors that are reported.

answered Dec 1, 2010 at 19:11

Ian

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1 Comment

KawaGreen Over a year ago

I discovered an error in the HTML: <select name="fk_keten_id"><br /> <b>Warning</b>: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in <b>/home/****/addLocatie.php</b> on line <b>56</b><br /> </select> I'm not working on localhost so I can't enable the errors

Jose Vega · Accepted Answer · 2010-12-01 19:12:20Z

0

This part of your code looks fine:

echo '<select name="fk_keten_id">';             
while($row = mysql_fetch_array($allKet))
{
  echo '<option value="' . 
       $row['pk_keten_id'] . '">' . $row['keten_naam'] . '</option>';
}
echo '</select>';

Your problem should lie in $allKet.

Do you need to give Keten() any arguments to initialize anything?

edited Dec 1, 2010 at 19:12

answered Dec 1, 2010 at 19:00

Jose Vega

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Comments

KawaGreen · Accepted Answer · 2010-12-01 19:28:53Z

0

Ok I got it

use: while($row = mysqli_fetch_assoc($allKet))

instead of: while($row = mysql_fetch_array($allKet))

Thank you all for your fast replies !

answered Dec 1, 2010 at 19:28

KawaGreen

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2 Comments

zanlok Over a year ago

it's nice to the community (and our reputations) to mark whichever answer helped you (the most) as "accepted" -- the green checkmark

KawaGreen Over a year ago

Yeah I know, but I need more points to do that :) I will do it in the future

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