form values

[sofia403]

hi i was wondering how i can display the values for a form that already in db. users submitted a form and then when they log in to their profile they can change the value in a form, but the original value would show the one that they first submitted. makes sense?

 

at the moment its set as blank, how would i display the option value to the one in the db on their profile? thanks

 

<form name="test" method="post">
	<select name="ApplicationStatus">
	<option value="'">------</option>
	<option value="In process">In process</option>
	<option value="Withdrawn">Withdrawn</option>
                </select>

[The Little Guy]

Like so:

 

<?php
echo '<form name="test" method="post">
	<select name="ApplicationStatus">
	<option value="\'">------</option>
	<option value="In process" '.(($row['column1']=='In process')?'selected="selected"':'').'>In process</option>
	<option value="Withdrawn" '.(($row['column1']=='Withdrawn')?'selected="selected"':'').'>Withdrawn</option>
                </select>';
?>

[sofia403]

i dont get it

 

i set as in example below but i just get blank as my first option, the value that user submitted in db is not displayed. what am i missing?

[mikesta707]

Can you post the code you are using? It's hard to give a definitive answer without seeing what you are working with

[sofia403]

ok here it is

 

<?php
include('config.php');
$uname=$_SESSION['username'];
if($_SESSION['username']){
$sql="SELECT ApplicationStatus FROM table1 WHERE uname='$uname'";
$result=mysql_query($sql);}
echo '<form name="test" method="post">
	<select name="ApplicationStatus">
	<option value="">------</option>
	<option value="In process" '.(($row['column1']=='In process')?'selected="selected"':'').'>In process</option>
	<option value="Withdrawn" '.(($row['column1']=='Withdrawn')?'selected="selected"':'').'>Withdrawn</option>
                </select>';
?>

[mikesta707]

Simple error.

 

you never define $row.

 

perhaps you mean to stick the following line after your $result=.. line

$row = mysql_fetch_array($result);

 

also, I don't know how your table is structured, but are you sure column1 is the correct column? Instead of column1 perhaps it should be ApplicationStatus

[sofia403]

hehe thanks! that worked!