PHP : Not able to show value retrieved from database into dropdown - php

Question

I have an Task creation - View - Edit page. Once I create the task and user wants to edit it. He clicks edit button. So the value gets populated according to id. All the value gets populated except for Dropdown.:
This are my dropdowns :
<b>Assignee: &nbsp &nbsp &nbsp &nbsp </b><select name = "assignee" value = <?php echo $assignee ?></select>
<b>Priority:</b><select name = "priority" value= "<?php echo $priority; ?>" id="priority"><option>Low</option><option>Medium </option><option>High</option></select>
<b>Status: </b><select name = "status" value= "<?php echo $status; ?>" ><option>Assigned</option><option>Yet to Start </option><option>In Progress</option><option>Completed</option><option>Blocked</option></select>
This is code for getting the values and showing in table and updating to database
<?PHP
function renderForm($id, $task, $comments, $assignee, $priority, $status, $dataum1, $dataum2, $error) {/connecttothedatabaseinclude ('configdb1.php');
// check if the form has been submitted. If it has, process the form and save it to the database
if (isset($_POST['submit'])) {
// confirm that the 'id' value is a valid integer before getting the form data
if (is_numeric($_POST['id'])) {
// get form data, making sure it is valid
$id = $_POST['id'];
$task = $_POST['task'];
$comments = $_POST['comments'];
$assignee = $_POST['assignee'];
$priority = $_POST['priority'];
$status = $_POST['status'];
$dataum1 = $_POST['dataum1'];
$dataum2 = $_POST['dataum2'];
// check that firstname/lastname fields are both filled in
if ($task == '' || $comments == '') {
// generate error message
$error = 'ERROR: Please fill in all required fields!';
//error, display form
renderForm($id, $task, $comments, $assignee, $priority, $status, $dataum1, $dataum2, $error);
} else {
// save the data to the database
mysql_query("UPDATE work SET task='$task', comments='$comments', assignee='$assignee', priority='$priority', status='$status', dataum1='$dataum1', dataum2='$dataum2' WHERE id='$id' ") or die(mysql_error());
// once saved, redirect back to the view page
header("Location: view.php");
}
} else {
// if the 'id' isn't valid, display an error
echo 'Error!';
}
} else
// if the form hasn't been submitted, get the data from the db and display the form
{
// get the 'id' value from the URL (if it exists), making sure that it is valid (checing that it is numeric/larger than 0)
if (isset($_GET['id']) && is_numeric($_GET['id']) && $_GET['id'] > 0) {
// query db
$id = $_GET['id'];
$result = mysql_query("SELECT * FROM work WHERE id=$id") or die(mysql_error());
$row = mysql_fetch_array($result);
// check that the 'id' matches up with a row in the databse
if ($row) {
// get data from db
// get data from db
$task = $row['assignee'];
$comments = $row['2'];
$assignee = $row['assignee'];
$priority = $row['priority'];
$status = $row['status'];
$dataum1 = $row['dataum1'];
$dataum2 = $row['dataum2'];
// show form
renderForm($id, $task, $comments, $assignee, $priority, $status, $dataum1, $dataum2, '');
} else
// if no match, display result
{
echo "No results!";
}
} else
// if the 'id' in the URL isn't valid, or if there is no 'id' value, display an error
{
echo 'Error!';
}
}
?>

Answer 1

The select element does not have a value attribute - the selected option has a selected attribute.
In other words, you want something like:
<select name = "priority" id="priority">
<option <?php if ($priority == 'Low') { echo 'selected="selected"'; } ?>>Low</option>
<option <?php if ($priority == 'Medium') { echo 'selected="selected"'; } ?>>Medium </option>
<option <?php if ($priority == 'High') { echo 'selected="selected"'; } ?>>High</option>
</select>