PHP MYSQL Changing the photo on my database - php

Question

I am very new to PHP;
ADD.PHP - I have a form that collects the following information 1. name 2. email 3. phone and picture
pictures are stored on a directory folder on my server and then the filename of that photo is stored on my sql table.
VIEW.PHP - all the data in mysql is being displayed in this page including the photo in tabular format including the id of every record. The id being display is a hyperlink in which when clicked you will be directed to a page wherein you can edit the record contents:
below is the code for my EDIT.PHP
<?php
// Connects to your Database
mysql_connect("localhost", "user1", "12345") or die(mysql_error()) ;
mysql_select_db("test") or die(mysql_error()) ;
// Check whether the value for jobnumber is transmitted
if (isset($_GET['id'])) {
// Put the value in a separate variable
$id = $_GET['id'];
// Query the database for the details of the chosen jobnumber
$result = mysql_query("select id, name, email,
phone, picture from employees where id = $id");
// Check result
// This shows the actual query sent to MySQL, and the error. Useful for debugging.
if (!$result) {
$message = "Invalid query: " . mysql_error() . "\n";
$message .= "Whole query: " . $query;
die($message);
}
// Attempting to print $result won't allow access to information in the resource
// One of the mysql result functions must be used
// See also mysql_result(), mysql_fetch_array(), mysql_fetch_row(),etc.
while ($row = mysql_fetch_array($result)) {
$name = $row['name'];
echo $name. "\n";
echo $row['email'] . "\n";
echo $row['phone'] . "\n";
echo "<img width=500px height=500px src=pics/" . $row['picture'].">" . "\n";
// form tag
echo '<form action="add2.php" method="POST">';
//display name
echo 'Name: <input type="text" name="name" value="';
echo $row['name'];
echo '"><br>';
//display email
echo 'email: <input type="text" name="email" value="';
echo $row['email'];
echo '"><br>';
//display phone
echo 'Phone: <input type="text" name="phone" value="';
echo $row['phone'];
echo '"><br>';
//display photo
echo 'Photo: <input type="text" name="photo" value="';
echo $row['picture'];
echo '"><br>';
echo '<input type="submit" value="Add">';
echo '</form>';
}
} else {
die("No valid data specified!");
}
?>
using this code, the test fields went well but the input box for the photo is blank and when i click the button the photo field in my database will be blank unless i uploaded a new photo?
how can the user change the existing photo? or retain the old photo if not being changed?

Answer 1

Assuming your employees.picture column stores the path to the image.
To upload a new photo, you're going to need to change a couple of things...
Your form needs to use the correct encoding type, ie
<form action="add2.php" method="post" enctype="multipart/form-data">
You also need to provide a "file" input element to accept the new photo, eg
<input type="file" name="photo">
To see if a new photo has been supplied, simply check (after detecting a valid POST request)
if ($_FILES['photo']['error'] == UPLOAD_ERR_OK) {
// file uploaded
}

Answer 2

see this link :http://www.w3schools.com/PHP/php_file_upload.asp
As to changing the existing photo, u should also save the name of the photo name in your db fom within the page add2.php
when any users wants to upload the photo for the second time, then a check should be made in add2.php to find whether a photo was previously uploaded. Then U can take a decision from the user from 2 yes, no buttons. If yes then UPDATE( not insert) the corresponding column in db table.
U can use jquery to take the decision and work accordingly. Any help needed with jquery?
IF it is for the first time that the uploading is gonna take place, then u can bypass the check process.
Clear?