PHP Image Not Coming Up [duplicate] - php

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This question already has an answer here:
Closed 11 years ago.
Possible Duplicate:
Why isn't my image showing up?
I have a weird problem here. I have this line of code where it works on one page but it doesnt on another. The PHP code is as follows:
PHP Page That Shows Image
<table border=1>
<tr>
<td align=center>EDIT</td>
</tr>
<tr>
<td>
<table>
<?
$id = $_GET['product_id'];
$result = mysql_query("SELECT * FROM products WHERE serial = '$id'");
$info = mysql_fetch_array($result);
?>
<form method="post" action="editsuccess.php">
<input type="hidden" name="id" value="<? echo "$info[name]"?>">
<table border='0' width=100%>
<tr>
<td>Name</td>
<td>
<input type="text" name="name"
size="20" value="<? echo "$info[name]"?>">
</td>
</tr>
<tr>
<td>Description</td>
<td>
<input type="text" name="name"
size="20" value="<? echo "$info[description]"?>">
</td>
</tr>
<tr>
<td>Price</td>
<td>
<input type="text" name="address" size="40"
value="<? echo "$info[price]"?>">
</td>
</tr>
<tr>
<td>Image</td>
<td>
<? echo'<img src="../getImage.php?id=' . $info['serial'] .'"/>'?>
</td>
</tr>
<tr>
<td align="right">
<input type="submit"
name="submit value" value="Update Product">
</td>
</tr>
</form>
</table>
</td>
</tr>
</table>
PHP Page That Doesnt Show Image
<?php
$id = $_GET['product_id'];
$query = mysql_query("SELECT * FROM products WHERE serial = '$id'")
or die(mysql_error());
while($info = mysql_fetch_array($query)) {
echo "";
$name = $info['name'];
$description = $info['description'];
$price = $info['price'];
$picture = $info['picture'];
}
?>
<form action="editsuccess.php?product_id=<?php echo $id; ?>" method="post">
Product ID:<br/>
<input type="text" value="<?php echo $id;?>" name="product_id" disabled/>
<br/>
Name:<br/>
<span id="sprytextfield1">
<input type="text" value="<?php echo $name;?>" name="name"/>
<span class="textfieldRequiredMsg">Enter Product Name</span></span><br/>
Description:<br/>
<span id="sprytextfield2">
<input type="text" value="<?php echo $description;?>" name="description"/>
<span class="textfieldRequiredMsg">Enter A Description</span></span><br/>
Price:<br/>
<span id="sprytextfield3">
<input type="text" value="<?php echo $price;?>" name="price"/>
<span class="textfieldRequiredMsg">Enter Price</span><span class="textfieldInvalidFormatMsg">Enter Numbers Only</span></span><br/>
Picture:<br/>
<?php echo '<img src="../getImage.php?id=' . $row['serial'] .'"/>'
?>
</br>
<input type="submit" value="Update Product"/>
</form>
The line of code i am talking about is this one:
<?php echo '<img src="../getImage.php?id=' . $row['serial'] .'"/>'
?>
Any ideas why it dont work???
-----EDIT--------
getImage.php code is as follows:
<?php
$host="localhost"; // Host name
$user="****"; // Mysql username
$passwd="****"; // Mysql password
$dbName="**********"; // Database name
// Connect to server and select databse.
mysql_connect("$host", "$user", "$passwd")or die("cannot connect");
mysql_select_db("$dbName")or die("cannot select DB");
$link = mysql_connect($host, $user, $passwd);
mysql_select_db($dbName);
$query = 'SELECT picture FROM products WHERE serial="' . $_GET['id'] . '"';
$result = mysql_query($query,$link);
$row = mysql_fetch_assoc($result);
header("Content-type: image/jpeg");
echo $row['picture'];
?>

Answer 1

First of all, try to avoid short tags (<?). Not every webserver is configured to understand them and it kind of conflicts with XML tags (which open with <?xml). So replace your <? with <?php to make sure your code always works on any webserver, regardless of it's configuration setting for short_open_tags.
Second, You're calling $row['serial'], but $row doesn't appear to be an array (at least it's not defined within the code you pasted here). Are you sure it shouldn't be $info['serial']?
But most importantly, whenever you allow user-input (like a $_GET) to determine your SQL query, always escape your code with mysql_real_escape_string, like this:
$result = mysql_query("SELECT * FROM products WHERE serial = '" . mysql_real_escape_string($id) . "'");
Or when you're sure that it's always an integer (e.g. if the field has INT datatype in your database), cast the value as an integer, like so:
$result = mysql_query("SELECT * FROM products WHERE serial = " . (int) $id);

Answer 2

You shooed to remove while cicluse
next code
while($info = mysql_fetch_array($query)) {
echo "";
$name = $info['name'];
$description = $info['description'];
$price = $info['price'];
$picture = $info['picture'];
}
?>
change just with
$info = mysql_fetch_array($query)
Try and tell us does is ok, and does is like you want. :)