dropdown list from mysql database

yebaws · Feb 11, 2009

I want to set an "initially selected" option for a drop down list to be the same as the data in a field from a database query. For example:

Code:

<select name="fruit" id="fruit">
      <option value="apples">apples</option>
	  <option value="pears">pears</option>
</select>

If I query the database and the value returned for the "fruit" field is "apples", then I want to write a drop down menu like this:

Code:

<select name="fruit" id="fruit">
      <option value="apples" selected>apples</option>
	  <option value="pears">pears</option>
</select>

So that you can see from the drop down what value is in the database, and change it if you want.

Any ideas?

Thanks.

vacunita · Feb 11, 2009

You'll need to dynamically create your drop dowwn
i.e:

Code:

<?PHP
$output.="";

$output.="<option value=apples ";

if($value_from_db=='apples'){
$output.="selected=selected>";
}

$output.="Apples</option>";

$output.="<option value=pears ";
if($value_from_db=='pears'){
$output.="selected=selected>";
}

$output.="Pears</option>";

...

?>

----------------------------------
Ignorance is not necessarily Bliss, case in point:
Unknown has caused an Unknown Error on Unknown and must be shutdown to prevent damage to Unknown.

yebaws · Feb 11, 2009

Thank you. That does the job.

I was hoping there might be a slightly less long winded way to do it as I have a bit more than apples and pears, but I guess not.

vacunita · Feb 11, 2009

If you are generating the drop down from something such as a database or even an array, you can just loop through the list of options and have a single if statement to generate the selected item.

----------------------------------
Ignorance is not necessarily Bliss, case in point:
Unknown has caused an Unknown Error on Unknown and must be shutdown to prevent damage to Unknown.

jpadie · Feb 13, 2009

i use this method of generating options

Code:

function getOptions($tableName, $value, $display, $where = '', $selected=''){
 $options = array();
 $where = empty($where) ? '' : 'where $where';
 $sql = "select $value, $display from $tableName $where";
 $result = mysql_query($sql);
 while ($row = mysql_fetch_array($result)){
  $_selected = ($row[0] == $selected) ? 'selected="selected" :'';
  $options[] = <<<HTML
 <option value="{$row[0]}" $_selected>{$row[1]}</option>
HTML;
 } //end while
 return implode("\r\n", $options);
}

and to use the function in your code

Code:

echo "<select name=\"somefield\">";
echo getOptions('fruits', 'id', 'fruitName');
echo "</select>";

you can use the fourth parameter to filter the options and fifth to make forms sticky.

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dropdown list from mysql database

yebaws

Technical User

vacunita

Programmer

yebaws

Technical User

vacunita

Programmer

jpadie

Technical User

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