0

im having problem with this 2 pages of code i have.the only problem is when im going to display it in another page called view.php

added.php

<?php

$type = $_POST['type'];
$size = $_POST['size'];
$quantity = $_POST['quantity'];
$prize = $_POST['prize'];
$file = $_FILES['image']['tmp_name'];

if (isset($_POST['submit']))
{
	include('sql.php');
	if (!isset($file)){
	
	echo "Please Select an Image";
	}
	else{
	
	$image = addslashes(file_get_contents($_FILES['image']['tmp_name']));
	$image_name = addslashes($_FILES['image']['name']);
	$image_size = getimagesize($_FILES['image']['tmp_name']);
	
	if(!$image_size){
	echo "Invalid File Type";}
	else{
		
	$query= "INSERT INTO tbcatalog VALUES('','$image_name','$image','$type','$size','$quantity','$prize')";	
	$result = @mysql_query($query);
	$lastid = mysql_insert_id();
	echo "Product Successfully Added<img src=view.php?id=$lastid>";
	mysql_close();
	
	}
		}
	}

?>

view.php

<html>
<head>
<link rel="icon" href="images/logo1.png" type="images/png	"> 
<meta content="text/html; charset=utf-8" http-equiv="Content-Type" />
<link rel="stylesheet" href="style.css" type="text/css">
</head>
<body bgcolor="white"><table align="center" border="1" width="100%" height="25%"><td><img src="images/images-2.jpeg" width="1020" height="290"></td></table>
<div class="style1" align="left">
<p class="title">Products</p>
<?php
// Script name: view.php
// Description: The display/view all records in the database

include('sql.php');
header("Content-type: image/jpeg");

$id = $_GET['id'];
$query = "select * FROM tbcatalog where id='".$id."'";
$result = @mysql_query($query);

mysql_close();
?>

<table class="table3" border="1" cellspacing="3" cellpadding="3">
    <tr>
    <th class="text2">#</th><th class="text2">ID</th><th class="text2">Product Name</th><th class="text2">Image</th><th class="text2">Product Type</th><th class="text2">Size</th><th class="text2">Stock</th><th class="text2">Prize</th><th class="text2">Task</th>
	</tr>
<?php
$i=1;
while ($row = mysql_fetch_assoc($result)){
$img = $row['image'];

    
    echo "<tr><td bgcolor=\"white\">".$i."</td>";
    echo "<br";
    echo "<tr><td bgcolor=\"white\">".$row['id']."</td>";
    echo "<br";
    echo "<tr><td bgcolor=\"white\">".$row['name']."</td>";
    echo "<br";
    echo "<tr><td bgcolor=\"white\">".$img."></td>";
    echo "<br";
    echo "<tr><td bgcolor=\"white\">".$row['type']."</td>";
    echo "<br";
    echo "<tr><td bgcolor=\"white\">".$row['size']."</td>";
    echo "<br";
    echo "<tr><td bgcolor=\"white\">".$row['quantity']."</td>";
    echo "<br";
    echo "<tr><td bgcolor=\"white\">".$row['prize']."</td>";
    echo "<br";
	echo "<tr><td bgcolor=\"white\"><a href='update.php?id=" . $row['id'] . "'><img alt=\"Edit\"  height=\"15\" src=\"icons/edit-icon.jpg\" width=\"15\"></a><img alt=\"Delete\"  height=\"15\" src=\"icons/delete-icon.jpg\" width=\"15\"><img alt=\"Edit\"  height=\"15\" src=\"icons/view-icon.jpg\" width=\"15\"></td>";
    echo "</tr>";
    echo "<br>";
    echo "<img src='view.php?id=".$id."'>";

	$i++;   

}
mysql_free_result($result);
?>
<h3 style="font-family:'Arial Black'">total products: <?php echo($i-1); ?> </h3> 


 
 </table>
</html>

when i execute the view.php its show a broken image

any help is much appreciated ty

3
Contributors
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Replies
4
Views
5 Years
Discussion Span
Last Post by kewl
0

It's giving you the `Broken Image` error because you aren't displaying just a image when calling the view.php file from other scripts.
Maybe have a function that returns the value of the image from the database and uses the data:// Uri, for example?

0

Replace the line 40

echo "<tr><td bgcolor=\"white\">".$img."></td>";

with

echo "<tr><td bgcolor=\"white\"><img src='foldername/$img'></td>";

and no need of this line

echo "<br";

if you are using <table>

if you need change it as

echo "<br>";

Edited by karthik_ppts: n/a

0

Replace the line 40

echo "<tr><td bgcolor=\"white\">".$img."></td>";

with

echo "<tr><td bgcolor=\"white\"><img src='foldername/$img'></td>";

and no need of this line

echo "<br";

if you are using <table>

if you need change it as

echo "<br>";

_thank you for this i'll try it later when i get home

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