display image

Dekpo

please i need help on this code.the code is made of an html which uploads image and a php script which displays the image and its size.my problem is from the php code;it doesn't displays the image after upload using the html .pls i ll appreciate an elaborate discussion hilie i show the code.

1.the html code

html>
<head>
<title>Upload your pic to our site!</title>
</head>
<body>
<form name="form1" method="post" action="check_image.php"
enctype="multipart/form-data">
<table border="0" cellpadding="5">
<tr>
<td>Image Title or Caption<br>
<em>Example: You talkin' to me?</em></td>
<td><input name="image_caption" type="text" id="item_caption" size="55"
maxlength="255"></td>
</tr>
<tr>
<td>Your Username</td>
<td><input name="image_username" type="text" id="image_username" size="15"
maxlength="255"></td>
</tr>
<td>Upload Image:</td>
<td><input name="image_filename" type="file" id="image_filename"></td>
</tr>
</table>
<br>
<em>Acceptable image formats include: GIF, JPG/JPEG, and PNG.</em>
<p align="center"><input type="submit" name="Submit" value="Submit">
 
<input type="reset" name="Submit2" value="Clear Form">
</p>
</form>
</body>
</html>

2.the php code

<?php

//connect to the database
$link = mysql_connect("localhost", "root", "password")
or die("Could not connect: " . mysql_error());
mysql_select_db("moviesite", $link)
or die (mysql_error());
//make variables available
$image_caption = $POST['image_caption'];
$image_username = $POST['image_username'];
$image_tempname = $_FILES['image_filename']['name'];
$today = date("Y-m-d");
//upload image and check for image type
//make sure to change your path to match your images directory

$ImageDir ="C:/wamp/www/phphr/Clips";
$ImageName = $ImageDir.$image_tempname;
if (move_uploaded_file($_FILES['image_filename']['tmp_name'],$ImageName)){

//get info about the image being uploaded

list($width, $height, $type, $attr) = getimagesize($ImageName);
switch ($type) {
case 1:
$ext = ".gif";
break;
case 2:
$ext = ".jpg";
break;
case 3:
$ext = ".png";
break;
default:
echo "Sorry, but the file you uploaded was not a GIF, JPG, or " .
"PNG file.<br>";
echo "Please hit your browser's 'back' button and try again.";
}
//insert info into image table
$insert = "INSERT INTO images
(image_caption, image_username, image_date)
VALUES
('$image_caption', '$image_username', '$today')";
$insertresults = mysql_query($insert)
or die(mysql_error());
$lastpicid = mysql_insert_id();
$newfilename = $ImageDir.$lastpicid.$ext;
rename($ImageName,$newfilename);
}

<html>
<head>
<title>Here is your pic!</title>
</head>
<body>
<h1>So how does it feel to be famous?</h1><br><br>
<p>Here is the picture you just uploaded to our servers:</p>
<img src="Clips/<?php echo $lastpicid.$ext; ?>" align="left">
<strong><?php echo $image_caption; ?></strong><br>
This image is a <?php echo $ext; ?> image.<br>
It is <?php echo $width; ?> pixels wide
and <?php echo $height; ?> pixels high.<br>
It was uploaded on <?php echo $today; ?>.
</body>
</html>

alisun

I would like to see some discussion on this. I'm trying to do something similar.

NogDog

Typically you'll need a separate PHP script (or separately invoked action in a framework's controller) whose only purpose is to output the image. Then you invoke it via the SRC attribute of an IMG tag in your HTML output, usually with a URL query string value that is used to query the DB to determine which image gets output.

<img src='/image_handler.php?id=<?php echo (int)$imgId; ?>' alt='' />

Then in this example, image_handler.php would query the DB based on $_GET['id'], and perhaps do a readfile() to output the applicable image, first using header() to set the correct Content-Type: header based on the image type info from the DB.

alisun

Thank you for your response.

I've been trying to digest this over the last couple of days.

This is still way over my head at this point.

I will continue working on it, but it's going to take some time.

NogDog

Let's try a fairly simple example:

image.php:

<?php
// for example, replace with your preferred DB connection stuff
require_once $_SERVER['DOCUMENT_ROOT'].'/includes/dbconnect.php';

if(!empty($_GET['id'])) {
  $sql = "SELECT file_name FROM images WHERE image_id = " . (int) $_GET['id'];
  if(($result = mysql_query($sql)) != false and mysql_num_rows() > 0) {
    $row = mysql_fetch_assoc($result);
    $filePath = $_SERVER['DOCUMENT_ROOT'].'/images/'.$row['file_name'];
    $size = getimagesize($filenPath);
    $fp = fopen($filename, "rb");
    if ($size and $fp) {
      header("Content-type: {$size['mime']}");
      fpassthru($fp);
      exit;
    }
  }
}
header("HTTP/1.0 404 Not Found");

HTML output of main script:

<?php
/* some processing to get appropriate image id from DB and store in variable $imgId happens here, then... */
?>
<img src='/image.php/id=<?php echo (int) $imgId;?>' alt='' />