how to populate drop down value?

hongmei

Hi,

I have a drop down box with static values. When user choose "Routine", then click submit button, I want the screen will dipslay "Routine" in the drop down. What should I code?

Here is part of my code:
But I only get the value displayed in the drop down. How can I get option selected not the value itself?

Thanks!

hongmei

<td colspan="1" width="180">
<select name="selInitPriority" size="1">
<option value="<?php
if (isset($POST['selInitPriority'])) {
echo htmlspecialchars($POST['selInitPriority']);
}?> "><?php echo $_POST['selInitPriority'] ?></option>
<option value=""></option>
<option value="routine">Routine</option>
<option value="urgent">Urgent</option>
<option value="emergency">Emergency</option>
</select>
</td>

John_Canyon

$option_array = array("option1","option2","option3","option4");

$select = "<select name=yourname>";
for($i=0;$i<count($option_array);$i++)
{
if($_POST[yourvalue] == $option_array[$i])
{
$select .= "<option value=".$option_array[$i]." selected>".$option_array[$i]."</opton>";
}
else
{
$select .= "<option value=".$option_array[$i].">".$option_array[$i]."</opton>";
}
$select .= "</select>";

}

hongmei

Hi, John:

Thank you very much for your kindly reponse!
But in your code:

$select .= "<option value=".$option_array[$i]." selected>".$option_array[$i]."</opton>";

The storing value and displaying value are the same. You used $option_array[$i] to refer them. They are usually different. For example, value is "urgent" and the displaying value is "Urgent".

How to correct refer th displaying value, not the real value?

hongmei

jmarco

$select .= "<option value=".$option_array[$i]." selected>".ucfirst($option_array[$i])."</opton>";

The function

ucfirst()

will upper case the first letter of a string.

http://www.php.net/manual/en/function.ucfirst.php

hongmei

Thank you for making me laugh!