How to access data from database and make an option value?

Question

I have this code

$result = mysql_query("SELECT name from room");
$data = mysql_fetch_array($result); 
$firstValue="";
while($data = mysql_fetch_array($result)){ 
    if($firstValue==""){ 
        $firstValue=$data['name'];
    }
    if(isset($_POST["occupant"])and trim($_POST["choice"])==$data['name']){
        echo '<option selected="selected" value="'.$data['name'].'" >'.$data['name'];
        echo '</option>';
    }else
        {
         echo '<option value="'.$data['name'].'" >'.$data['name'];
         echo '</option>';
         }
    }

What i want here is to show all data from my database but when i open it in browser shows only the second data down to the last data. And i wonder why the first data is missing. Can anybody knows my mistake here?

Also, note, you should switch to the mysqli_ or PDO extension as the mysql_ extension is going to be deprecated as of PHP 5.5.0. It is an easy change and good to do it as you start out rather than later on. :-) — brechmos
– brechmos, Commented Jun 13, 2014 at 13:42
I don't really follow what 'second data' is. You mean you want to skip the first record altogether? Just filter it out in your SQL. use the LIMIT clause: LIMIT 999, 1. Or use WHERE id > 1. — Richard Testani
– Richard Testani, Commented Jun 13, 2014 at 13:46
how to switch to mysqli? @Richard Testani, i don't want to skip the first, what i want is to show all data.. — Jay-Ar Aclan
– Jay-Ar Aclan, Commented Jun 13, 2014 at 14:05

John Conde · Accepted Answer · 2014-06-13 13:38:58Z

2

Get rid of the first:

$data = mysql_fetch_array($result);

It is popping the first record off of your result set.

$result = mysql_query("SELECT name from room");
$firstValue="";
while($data = mysql_fetch_array($result)){

answered Jun 13, 2014 at 13:38

John Conde

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4 Comments

Jay-Ar Aclan Over a year ago

i did this many times but still the first data is missing.. how to set off?

John Conde Over a year ago

I did. I even showed you how to fix it. What part of my answer do you not understand that I might need to clarify for you?

Jay-Ar Aclan Over a year ago

do i have to erase this block of code? $data = mysql_fetch_array($result);

John Conde Over a year ago

Yes. That is exactly what you need to do.

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