dynamic dropdown list

tbarca

Hi,

I am trying to generate a dropdown box with values from a mysql table with the below text, but for som reason it doesn't return any values in the list. could it be because of the syntax in line 5? ($myrow["$column"])

thanks,
Tim

1 function Dropdown($selectname,$table,$column) {
2 echo "<select name='".$selectname."'>";
3 $result = mysql_query("SELECT $column FROM $table ORDER by $column ASC");
4 while ($myrow = mysql_fetch_array($result)) {
5 printf("<option value='%s'>%s</option>\n", $ myrow["$column"],$myrow["$column"]);
6 }
7 echo "</select>";
8 }
9 echo Dropdown("taskstatus","dd_status_task","stt_value");

rem

Well, I think that's just because you echo() the returning value of the function Dropdown, but it doesn't return anything.

You should change the functions like this:

function Dropdown($selectname,$table,$column) {
$sReturn = "<select name='".$selectname."'>";
$result = mysql_query("SELECT $column FROM $table ORDER by $column ASC");

while ($myrow = mysql_fetch_array($result)) {
$sReturn .= "<option value=\"" . $myrow["$column"]. "\"";
$sReturn .= $myrow["$column"]. "</option>\n";
}

$sReturn .= "</select>"; 
return $sReturn;

}

Regards

ReM

tbarca

i tried that but it still returns nothing - am i supposed to use
echo Dropdown(value1,value2,value3)
to return the values, or is echo() not the correct function to display it?

thanks,

Tim

rossvri

i usually first assign the array to variables just to make them easier for me to understand...such as this

while ($row = mysql_fetch_array($result))
{
$variable = $row[0];

echo "<option value = '$variable'>$variable</option>

}

also, i noticed that you have a space after your $ in row 5. that could definetely cause an error.

rem

nothing in zend's error log ?