MySQL PHP filter array so only non empty values are filled into a form select menu

Question

I believe its quite a trivial task but I couldn't figure it out yet. I have a mysql table with multiple rows and columns. Each column should be used as a category for a dropdown menu. However some of those columns are shorter than the other ones. I have it currently implemented like this:

<select name="exhaust">
<option value="<? echo "$exhaust"; ?>" selected><? echo "$exhaust"; ?></option>
<?
//connect to mysql
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database"); 
$query = "SELECT * FROM tuning_parts";  
$query = mysql_query($query);
//ausgabe
while($db = mysql_fetch_array($query)){
$phrase = "<option value=\"".$db['exhaust']."\">".$db['exhaust']."</option>";

 echo($phrase);

  };
  ?>

</select>

However, this give me sometimes very long dropdown lists with a lot of empty values. I've tried to play around with array_filter() but I always got empty results.

I would like to filter out the empty fields so the dropdown menu only shows actual values.

Shakti Singh · Accepted Answer · 2012-03-05 07:49:11Z

2

The better way is don't select those records which are empty instead picking up them from database and preventing in code level.

Change your query to

$query = "SELECT * FROM tuning_parts 
          WHERE exhaust IS NOT NULL 
               AND exhaust !='' ";

answered Mar 5, 2012 at 7:49

Shakti Singh

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Comments

Sudhir Bastakoti · Accepted Answer · 2012-03-05 07:48:22Z

0

Well, you could do:


//inside your while loop
if(!empty($db['exhaust'])) {
  $phrase = "<option value=\"".$db['exhaust']."\">".$db['exhaust']."</option>";

}

answered Mar 5, 2012 at 7:48

Sudhir Bastakoti

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Comments

Oyeme · Accepted Answer · 2012-03-05 07:50:12Z

0

Just smple use 'if' condition

   if($db['exhaust']  != ''){
        $phrase = "<option value=\"".$db['exhaust']."\">".$db['exhaust']."</option>";
    }

answered Mar 5, 2012 at 7:50

Oyeme

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