Display image in MYSQL

[acellec]

hello i added a field in my table for images with field type long blob...how can i display it in my php code.. ihave this in my code:

Row 6 is the field for images please help....

i have added this code to display but it doenst show any image ..thanks
echo "<td align='right'><img src=\"$Row[6]\"alt=\"\"/></td>";

<?

$TableName = "automobileparts";
$SQLstring = "SELECT * FROM automobileparts";

require_once("dbconnect.php");



        
echo "<table width='100%' border='1'>";
$Row = $QueryResult->fetch_row();
do {
             echo "<tr><td>{$Row[0]}</td>";
        
        echo "<td align='right'>{$Row[1]}</td>";
        echo "<td align='right'>{$Row[2]}</td>";
        echo "<td align='right'>{$Row[3]}</td>";
        echo "<td align='right'>{$Row[4]}</td>";
        echo "<td align='right'>{$Row[5]}</td>"; 
        echo "<td align='right'><img src=\"$Row[6]\"alt=\"\"/></td>"; 
        echo "<td align='right'><a href=\"Mail2.php?id={$Row[0]}\">Order</a></td>";   
                       
                   echo "</tr>";
        $Row = $QueryResult->fetch_row();
} while ($Row);
echo "</table>";



$DBConnect->close();


?>

[garrettroyce]

You have to use the header() function to set the content-type as image/[your image type]

Since you want to use the image in a page that has more than just an image, you have to actually use two different scripts. Then, you will access your images like this:

<img src="get_image.php?image=$image_number" />

$image_number should be a unique id for your image. If your table isn't set up this way, add a row, integer type, unsigned, auto increment, primary key.

Then, getting the image is as easy as this: http://bytes.com/groups/php/1655-php...blobs#post4918

[acellec]

You have to use the header() function to set the content-type as image/[your image type]

Since you want to use the image in a page that has more than just an image, you have to actually use two different scripts. Then, you will access your images like this:

<img src="get_image.php?image=$image_number" />

$image_number should be a unique id for your image. If your table isn't set up this way, add a row, integer type, unsigned, auto increment, primary key.

Then, getting the image is as easy as this: http://bytes.com/groups/php/1655-php...blobs#post4918

it doest display can you check my code

<?

$TableName = "automobileparts";
$SQLstring = "SELECT * FROM automobileparts";

require_once("dbconnect.php");



header ("Content-type: image/jpeg");        
echo "<table width='100%' border='1'>";
$Row = $QueryResult->fetch_row();
do {
                 
        echo "<tr><td>{$Row[0]}</td>";
        echo "<td align='right'>{$Row[1]}</td>";
        echo "<td align='right'>{$Row[2]}</td>";
        echo "<td align='right'>{$Row[3]}</td>";
        echo "<td align='right'>{$Row[4]}</td>";
        echo "<td align='right'>{$Row[5]}</td>";
        echo "<td align='right'><img src="get_image.php?image=$Row[1]" /></td>";
        echo "<td align='right'></td>"; 
        echo "<td align='right'><a href=\"Mail2.php?id={$Row[0]}\">Order</a></td>";   
                       
                   echo "</tr>";
        $Row = $QueryResult->fetch_row();
} while ($Row);
echo "</table>";



$DBConnect->close();


?>

[Twinkie]

"header ("Content-type: image/jpeg");" should not be in the script printing the html, it should be in the script printing the image. Setting the header as an image ensures that the image data printed in get_image.php displays as an image rather than a bunch of incomprehensible characters.

[acellec]

"header ("Content-type: image/jpeg");" should not be in the script printing the html, it should be in the script printing the image. Setting the header as an image ensures that the image data printed in get_image.php displays as an image rather than a bunch of incomprehensible characters.

is it like this then ? still doesnt display..can you show me please

echo "<table width='100%' border='1'>";
$Row = $QueryResult->fetch_row();
do {
                 
        echo "<tr><td>{$Row[0]}</td>";
        echo "<td align='right'>{$Row[1]}</td>";
        echo "<td align='right'>{$Row[2]}</td>";
        echo "<td align='right'>{$Row[3]}</td>";
        echo "<td align='right'>{$Row[4]}</td>";
        echo "<td align='right'>{$Row[5]}</td>";
        
        header("Content-type: image/jpeg");
        print $Row[6];
        echo "<td align='right'></td>"; 
        echo "<td align='right'><a href=\"Mail2.php?id={$Row[0]}\">Order</a></td>";   
                       
                   echo "</tr>";
        $Row = $QueryResult->fetch_row();
} while ($Row);
echo "</table>"; 

[xav0989]

no, what you wrote will not work, since you a outputing both an image and some html. What Twikie meant was that the header('content-type: image/jpeg') should be in the get_image.php file. Also, instead of doing a do...while loop, you should do a while loop. Way better, if, for example, there are no results. And it will save you a couple lines. Oh and you should seriously consider using names instead of numerical indices. It's way clearer.

<?php
//display car info.php

$TableName = "automobileparts";
$SQLstring = "SELECT * FROM automobileparts";

require_once("dbconnect.php");

     
echo "<table width='100%' border='1'>";
while ($Row= $QueryResult->fetch_row()) {
                 
        echo "<tr><td>{$Row[0]}</td>";
        echo "<td align='right'>{$Row[1]}</td>";
        echo "<td align='right'>{$Row[2]}</td>";
        echo "<td align='right'>{$Row[3]}</td>";
        echo "<td align='right'>{$Row[4]}</td>";
        echo "<td align='right'>{$Row[5]}</td>";
        echo "<td align='right'><img src=\"get_image.php?image={$Row[1]}\" /></td>";
        echo "<td align='right'></td>"; 
        echo "<td align='right'><a href=\"Mail2.php?id={$Row[0]}\">Order</a></td>";             
        echo "</tr>";
}

echo "</table>";

$DBConnect->close();
?>

Next, in the get_image.php file:

<?php
//get_image.php

// you need to sanitize it yourself
$image_id = $_GET['image'];

$TableName = "automobileparts";
$SQLstring = "SELECT * FROM automobileparts WHERE id = $image_id LIMIT 1";

require_once("dbconnect.php");

header('Content-Type: image/jpeg');
$Row= $QueryResult->fetch_row()
echo $Row[6];
$DBConnect->close();
?>

[acellec]

no, what you wrote will not work, since you a outputing both an image and some html. What Twikie meant was that the header('content-type: image/jpeg') should be in the get_image.php file. Also, instead of doing a do...while loop, you should do a while loop. Way better, if, for example, there are no results. And it will save you a couple lines. Oh and you should seriously consider using names instead of numerical indices. It's way clearer.

<?php
//display car info.php

$TableName = "automobileparts";
$SQLstring = "SELECT * FROM automobileparts";

require_once("dbconnect.php");

     
echo "<table width='100%' border='1'>";
while ($Row= $QueryResult->fetch_row()) {
                 
        echo "<tr><td>{$Row[0]}</td>";
        echo "<td align='right'>{$Row[1]}</td>";
        echo "<td align='right'>{$Row[2]}</td>";
        echo "<td align='right'>{$Row[3]}</td>";
        echo "<td align='right'>{$Row[4]}</td>";
        echo "<td align='right'>{$Row[5]}</td>";
        echo "<td align='right'><img src=\"get_image.php?image={$Row[1]}\" /></td>";
        echo "<td align='right'></td>"; 
        echo "<td align='right'><a href=\"Mail2.php?id={$Row[0]}\">Order</a></td>";             
        echo "</tr>";
}

echo "</table>";

$DBConnect->close();
?>

Next, in the get_image.php file:

<?php
//get_image.php

// you need to sanitize it yourself
$image_id = $_GET['image'];

$TableName = "automobileparts";
$SQLstring = "SELECT * FROM automobileparts WHERE id = $image_id LIMIT 1";

require_once("dbconnect.php");

header('Content-Type: image/jpeg');
$Row= $QueryResult->fetch_row()
echo $Row[6];
$DBConnect->close();
?>

from the car info.php i just change this part to {$Row[0]}...since it's the part number id for the pictures

echo "<td align='right'><img src=\"get_image.php?image={$Row[1]}\" /></td>"; 

and from the get_image.php

i change this part id to PartNo.

$SQLstring = "SELECT * FROM automobileparts WHERE id = $image_id LIMIT 1"; 

i dont have errors but the picture still doesnt display..can you help me?
Edit:
myabe something wrong when i add a field image in mysql? what do you think?

[Twinkie]

What displays when you follow the direct link to the php image file?

[acellec]

What displays when you follow the direct link to the php image file?

it displays characters please help

[xav0989]

what kind of characters. There are loads of characters. Copy them here or give us some link that we can chew on...