How to make form values remain in form - php

Question

I'm developing a simple PHP website. I have a form for updating values. When that form is accessed via PHP(website) , It should have previous values, But it hasn't. How to do that ?
Because, Otherwise, It shows all fields empty and when one or two are updated, Other gets updated as empty. Please help !
<?php
require_once('../includes/config.php');
require_once('../includes/functions.php');
require_once('../includes/session.php');
require_once('../includes/database.php');
require_once('../includes/user.php');
require_once('../includes/photograph.php');
if (!$session->is_logged_in()) { redirect_to("login.php"); }
?>
<?php
// START FORM PROCESSING
if (isset($_POST['submit'])) { // Form has been submitted.
$errors = array();
// perform validations on the form data
$required_fields = array('lives_in', 'belongs_to', 'college', 'works_at', 'grade', 'grade2', 'hobbies', 'zodiac', 'phone_no', 'facebook', 'company', 'bio');
$errors = array_merge($errors, check_required_fields($required_fields, $_POST));
$lives_in = trim($database->escape_value($_POST['lives_in']));
$belongs_to = trim($database->escape_value($_POST['belongs_to']));
$college = trim($database->escape_value($_POST['college']));
$works_at = trim($database->escape_value($_POST['works_at']));
$grade = trim($database->escape_value($_POST['grade']));
$grade2 = trim($database->escape_value($_POST['grade2']));
$hobbies = trim($database->escape_value($_POST['hobbies']));
$zodiac = trim($database->escape_value($_POST['zodiac']));
$phone_no = trim($database->escape_value($_POST['phone_no']));
$facebook = trim($database->escape_value($_POST['facebook']));
$company = trim($database->escape_value($_POST['company']));
$bio = trim($database->escape_value($_POST['bio']));
if (empty($errors) ) {
$query = "UPDATE users
SET lives_in = '{$lives_in}',
belongs_to = '{$belongs_to}',
college = '{$college}' ,
works_at = '{$works_at}',
grade = '{$grade}',
grade2 = '{$grade2}',
hobbies = '{$hobbies}',
zodiac = '{$zodiac}',
phone_no = '{$phone_no}',
facebook = '{$facebook}',
company = '{$company}',
bio = '{$bio}'
WHERE id = 23
LIMIT 1";
$result = mysql_query($query, $database->connection);
if ($result) {
$message = "Account Updated";
} else {
$message = mysql_error();
}
} else {
if (count($errors) == 1) {
$message = mysql_error();;
} else {
$message = "There were " . count($errors) . " errors in the form.";
}
}
} else { // Form has not been submitted.
$lives_in = "";
$belongs_to = "";
$college = "";
$works_at = "";
$grade = "";
$grade2 = "";
$hobbies = "";
$zodiac = "";
$phone_no = "";
$facebook = "";
$company = "";
$bio = "";
}
?>

Answer 1

It's very easy, just print out the previous values in the form when you create it with php.
For example:
<input type="text" name="field_name" value="<?php echo $previous_value;?>" />
Or, if it's a select, let's say:
<select name="field_name">
<?php foreach($posible_values as $value){ ?>
<option value="<?php echo $value;?>"
<?php echo ($value==$previous_value)?'selected="selected"':'';?> >
<?echo $value?>
</option>
<?php } ?>
</select>