Undefined Index on PHP Drop Down using selected

Question

I am getting an Undefined index: for every ID which is over a thousand. I have tried everything to get the error to disappear and am hoping someone would be able to help. Please and Thanks in advance.

$bamid is what is called from the page and it works. The error comes from this statement "$selected[$id]" which there are no errors if removed. Adding double or single quotes does not work.

<?
$sql="SELECT * users ORDER BY name ASC";
$result=mysql_query($sql);
$options = "";
while ($row=mysql_fetch_array($result))
{
$id = $row['id'];
$name = $row['name'];
$selected = array($bamid => "selected");
$options.="<option value = \"$id\" $selected[$id]>".$name.'</option>';
}
?>
<select name = "bamid">
<option>-------</option>
<? echo $options?>
</SELECT>

Sample Output from the Error Log:

[Sun Jun 30 01:46:55 2013] [warn] [client xx.xx.xx.xx] mod_fcgid: stderr: PHP Notice:  Undefined index: 572 in xxxxxxxxxxxxxxx on line 173

[Sun Jun 30 01:46:55 2013] [warn] [client xx.xx.xx.xx] mod_fcgid: stderr: PHP Notice:  Undefined index: 833 in xxxxxxxxxxxxxxx on line 173

[Sun Jun 30 01:46:55 2013] [warn] [client xx.xx.xx.xx] mod_fcgid: stderr: PHP Notice:  Undefined index: 698 in xxxxxxxxxxxxxxx on line 173

[Sun Jun 30 01:46:55 2013] [warn] [client xx.xx.xx.xx] mod_fcgid: stderr: PHP Notice:  Undefined index: 666 in xxxxxxxxxxxxxxx on line 173

[Sun Jun 30 01:46:55 2013] [warn] [client xx.xx.xx.xx] mod_fcgid: stderr: PHP Notice:  Undefined index: 546 in xxxxxxxxxxxxxxx on line 173

[Sun Jun 30 01:46:55 2013] [warn] [client xx.xx.xx.xx] mod_fcgid: stderr: PHP Notice:  Undefined index: 688 in xxxxxxxxxxxxxxx on line 173

[Sun Jun 30 01:46:55 2013] [warn] [client xx.xx.xx.xx] mod_fcgid: stderr: PHP Notice:  Undefined index: 834 in xxxxxxxxxxxxxxx on line 173

[Sun Jun 30 01:46:55 2013] [warn] [client xx.xx.xx.xx] mod_fcgid: stderr: PHP Notice:  Undefined index: 312 in xxxxxxxxxxxxxxx on line 173

[Sun Jun 30 01:46:55 2013] [warn] [client xx.xx.xx.xx] mod_fcgid: stderr: PHP Notice:  Undefined index: 650 in xxxxxxxxxxxxxxx on line 173

[Sun Jun 30 01:46:55 2013] [warn] [client xx.xx.xx.xx] mod_fcgid: stderr: PHP Notice:  Undefined index: 1109 in xxxxxxxxxxxxxxx on line 173

[Sun Jun 30 01:46:55 2013] [warn] [client xx.xx.xx.xx] mod_fcgid: stderr: PHP Notice:  Undefined index: 430 in xxxxxxxxxxxxxxx on line 173

` echo $options?>` is missing a `;`, also what is on line 173 of the file in question on the error ? — Prix, Jun 30 '13 at 05:57
There is no `$bamid` on your code is it coming from a form ?? or did u mean `$id` there and wrote it wrongly — Prix, Jun 30 '13 at 05:59
$bamid comes from an echo. If the page contains "$bamid = 25", then select 25 from the drop down list, which it works. 25 is selected and displays, but the the undefined index is is for every "id" number. — user2535904, Jun 30 '13 at 06:03
That's because `$bamid` is unset on your while so it will be undefined for each row it prints. — Prix, Jun 30 '13 at 06:04

LSerni · Accepted Answer · 2013-06-30T06:12:51.030

Your code attempts to add a "selected" if the current $id equals $bamid, but what it does in all other cases? It doesn't know, so it throws a warning.

It still works, because undefined strings are rendered as empty strings in PHP, but it would be better to state this explicitly.

<?
// BTW, this opening tag would be better as <?php

// #include standard recommendation on moving to PDO instead of deprecated MySQL
// support

while ($row=mysql_fetch_array($result))
{
    $id = $row['id'];
    $name = $row['name'];
    $selected = array($bamid => "selected");
    $options.="<option value = \"$id\" $selected[$id]>".$name.'</option>';
}

You can do it like this:

$selected = ($id == $bamid) ? 'selected="selected"' : '';
$options.="<option value=\"$id\" $selected>$name</option>";

or, maybe making for more readable code at the expense of duplication:

if ($id == $bamid) {
    $options.= <<<OPTION_SELECTED

<option value="{$id}" selected="selected">{$name}</option>

OPTION_SELECTED;
} else {
    $options.= <<<OPTION_NOT_SELECTED

<option value="{$id}">{$name}</option>

OPTION_NOT_SELECTED;
}

If you do this a lot, you might want to write a utility function that takes an array of keys and values and zero or more selected keys/values, and returns the HTML for a combo box list of options. This would lead to

$optarr = array();
while ($row=mysql_fetch_array($result))
{
    $id   = $row['id'];
    $name = $row['name'];

    $optarr[$id] = $name;
}
mysql_free_result($result);

$options = htmlOptions($optarr, $bamid);

Thank you very much. It worked. No more errors. I agree, I am working on learning and changing to MySQLi soon. I plan on starting that later on this year. I also added the php after the question mark. My admin section is 8 years old and running PHP 5.5. Tons of errors I am trying to get fixed first. — user2535904, Jun 30 '13 at 06:17

score 0 · Answer 2 · answered Jun 30 '13 at 06:05

0

something like

$options.="<option value = \"$id\"";
if($bamid == $id) $options.= " selected";
$options.=">".$name.'</option>';

answered Jun 30 '13 at 06:05

Ian Kenney

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Undefined Index on PHP Drop Down using selected

2 Answers2