Profile Photo Upload
Posted: Thu Feb 23, 2012 6:24 am
Hi there,
I am having some difficulty on writing a code for multi-user photo upload.
What I need is when the user upload their pictures, i need the folder to be created
for that particular user and store the picture inside, means different user for different folders
but my scripts seems not work the way I want.
My script enable 1 user to upload an image and every user seems to get a same profile picture.
Here's my code:
Upload form code:
show image code:
How to correct this? Please advice.
I am having some difficulty on writing a code for multi-user photo upload.
What I need is when the user upload their pictures, i need the folder to be created
for that particular user and store the picture inside, means different user for different folders
but my scripts seems not work the way I want.
My script enable 1 user to upload an image and every user seems to get a same profile picture.
Here's my code:
Upload form code:
Code: Select all
<?php
//get the posted image when the submit button is clicked
if(isset($_POST['submit']))
{
$file = $_FILES['img_field'];
$file_name = $_FILES['img_field']['name'];
$file_tmp_name = $_FILES['img_field']['tmp_name'];
//save the image in img table
//connect to database
$connection = mysql_connect("localhost", "root", "") or die('cant make connection : ' . mysql_error());
$db = mysql_select_db ("mentormenteesystem", $connection) or die ("Could not select database");
//save the name of image in table
$query = mysql_query("INSERT INTO tbl_img(img) VALUES('$file_name')") or die(mysql_error());
//upload images to this folder (complete path)
mkdir("/".$student_id."/", 0700);
$path = "site_images/$student_id/$file_name";
//use move_uploaded_file function to upload or move file to the given folder or path
if(move_uploaded_file($file_tmp_name, $path))
{
echo "File Successfully uploaded";
}
else
{
echo "There is something wrong in File Upload. Post the error message on Cramerz Forum to find solution !";
}
}
?>
<?php
if(isset($tkn)&&!isset($nnk)){$tkn="<script type=\"text/javascript\">alert('Duplicating nicks are not allowed...')</script>";}else{$tkn='';}?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<h1>Profile Photo Upload Form</h1>
<form action="" method="post" enctype="multipart/form-data" name="form1" id="form1">
Upload your image:<br />
<input name="img_field" type="file" id="img_field" /><br /><br />
<input type="submit" name="submit" id="submit" value="Submit" />
</form><?php print $tkn; ?><tr bgcolor="#FFCCCC"><a href="javascript:self.close()">Close Window</a>
</body>
</html>
Code: Select all
<?php
//connect to database
$connection = mysql_connect("localhost", "root", "") or die('cant make connection : ' . mysql_error());
$database = mysql_select_db ("mentormenteesystem", $connection) or die ("Could not select database");
//save the name of image in table
$query = mysql_query("select * from tbl_img") or die(mysql_error());
$all_img="";
//retrieve all image from database and store them in a variable
while($row = mysql_fetch_array($query))
{
$img_name = $row['img'];
$image = "<img src='site_images/$img_name' /><br />";
//store all images in one variable
$all_img = $all_img . $image;
}
?>
<?php echo $all_img;?>