Getting a value in a dropdown box when updating a database value

Question

I am having a problem with updating some values in my database. When I want to change a value using a drop down box, it automatically puts in the first value in the dropdown menu. What I want is to get the value that is already set in the database.

Here is my code:

<select name="vrijwilligerID">
<?php           
    $vrijwilligerID = $_POST["vrijwilligerID"];
    $query = "SELECT voornaam, achternaam, vrijwilligerID FROM vrijwilliger;";
    $result = mysql_query($query);
    while($row=mysql_fetch_array($result)){
        echo "<option value=".$row['vrijwilligerID'].">".$row["voornaam"]." ".$row["achternaam"]."</option>";
    }
?>              
</select>

Does anyone know how to get this right? Thank you in advance.

What determines the selected value from the dropdownbox?

Chris Moutray
– Chris Moutray

2012-06-27 16:20:13 +00:00
Commented Jun 27, 2012 at 16:20 — Chris Moutray
– Chris Moutray, Commented Jun 27, 2012 at 16:20

Sudhakar B · Accepted Answer · 2012-06-27 16:18:19Z

2

Based on the selected item you have to add selected attribute to the option.

try this

<select name="vrijwilligerID">
                <?php           
                    $vrijwilligerID = $_POST["vrijwilligerID"];

                    $query ="   SELECT voornaam, achternaam, vrijwilligerID
                                FROM vrijwilliger;";

                    $result = mysql_query($query);

                    while($row=mysql_fetch_array($result)){
                      if($row["vrijwilligerID"]==$vrijwilligerID)
                           echo "<option value=".$row['vrijwilligerID']." selected>".$row["voornaam"]." ".$row["achternaam"]."</option>";

                     else
                          echo "<option   value=".$row['vrijwilligerID'].">".$row["voornaam"]." ".$row["achternaam"]."</option>";
                    }
                    ?>              
        </select>

answered Jun 27, 2012 at 16:18

Sudhakar B

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1 Comment

Rickkert Over a year ago

Doing this actually doesn't make any difference. It just gives the same output. So somehow it doesn't recognize the ID of something?

Amelia · Accepted Answer · 2012-06-27 16:41:59Z

Its easy enough to resolve that using php, but using mysqls IF() function you can have the selected option directly. Using mysqls concat() you could get your desired result as well.

(ofcourse we're not using any postsvars in our scripts unsanitized are we ;) )

$iVrijwilligerid = filter_input(INPUT_POST, 'vrijwilligerID', FILTER_VALIDATE_INT, array("options"=> array("min_range"=>0, "max_range"=>999))) ;

if($iVrijwilligerid)
{
  $sQry = <<<QRY
      SELECT
        CONCAT('<option value="', vrijwilligerID, '"',IF((SELECT vrijwilligerID FROM vrijwilliger WHERE vrijwilligerID=$iVrijwilligerid), ' selected="selected"', '' ),'>',voornaam, ,achternaam,'</option>') AS optItem
      FROM
        vrijwilliger
      WHERE
        your clause here
QRY;
  $oResult = mysql_query( $sQry );
  while($aRow=mysql_fetch_assoc( $oResult ))
  {
    echo $aRow['optItem'];
  }
}
else
{
  // nada
}

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