Echoing my BLOB image!?

[suess0r]

Hi there,

 

So i have the setup on the format I want to print all my result sets out, but I still don't know how to print the BLOB image?! I have a mediumblob setup and uploaded the images but i'm just not sure how to print them out. Here's what I gotz so far..

 

*Note: Where the "images/2_p4.jpg" is where I want to print the image! And it should be stored in $img

 

    $result_set = mysql_query($sql) or die("$sql FAILED because

  ".mysql_error());

 

  if (mysql_num_rows($result_set) > 0){

 

for ($i = 0; $i < mysql_num_rows($result_set); $i++)

{

$ename = mysql_result($result_set, $i, "Event_Name");

$date = mysql_result($result_set, $i, "Event_Date");

$vname = mysql_result($result_set, $i, "Venue_Name");

$description = mysql_result($result_set, $i, "Event_Description");

$cost = mysql_result($result_set, $i, "Event_Cost");

                $img = mysql_result($result_set, $i, "Thumb");

 

 

// on left

echo' <div style="margin:9 15 0 20px "><a href="#"><img src="images/2_p4.jpg" align="left" style="margin-right:15px " alt="" border="0"></a><span style="color: #00BCD4"><br style="line-height:9px">

                                              <span class="style4">'.$ename.' at '.$vname.'</span></span><br>

'.$description.'... <br><span class="style8">VIP Exclusive Cost of: '.$cost.'</span><br>

  <span class="style6" style="float:right; margin: 0 20px 0 0"><a href="#"><strong>read more</strong></a></span></div><br>

    <br style="line-height:11px"><img src="images/1_line.gif" style="margin-left:55px " alt="" border="0" width="425"><br><br style="line-height:10px">';

 

}

} else {

echo 'Sorry, no results were found...';

}

?>

 

Appreciate any help!

[suess0r]

anyone have any ideas about printing the blob images in the database?

[per1os]

You are going to need to make a separate file to retrieve the images.

 

IE:

get_image.php?imageid=1

 

//getimage.php

<?php
if (isset($_GET['imageid'])) {
    $qu = mysql_query("SELECT thumb FROM table_name WHERE imageid = " . $imageid) or die(mysql_error());
    $image = mysql_fetch_array($qu);

    header("Content-type: image/jpeg"); // unsure of what this should really be
    echo $image['thumb'];
}
?>

 

And than replace the images/2_p4.jpg with getimage?imageid=3

 

[dbillings]

Your best bet is to have a seperate script that will build your image and link to it via a hyperlink. You need to include a header with the correct MIME type as well which I don't see in your script.

 

<?php

// Create whatever filename is convenient for you I'll pick image.php.
// This will need to be saved as a seperate file
// this will generate your images when you link to it.
// I assume you are only using jpeg images otherwise 
// you will need a bit more code.

IF(isset($_GET['id'])){

// Set your mysql connect code here.

$id= $_GET['id'];
$query="SELECT Thumb FROM img where id='$id'";
$result= mysql_query($query)or die('Error: '.mysql_error());
$image= mysql_fetch_assoc($result);

//header("Content-type: image/pjpeg");


echo $image['Thumb'];





mysql_close();


echo $image;


}


?> 

 

Next you'll need to change your other script to retrieve a unique identifier for the image instead of the blob of the image itself. In this script I use the images id if you have one setup in mysql. Best practice is to have a seperate table for images but to each their own.

 

<?php

  $result_set = mysql_query($sql) or die("$sql FAILED because
  ".mysql_error());
  
  if (mysql_num_rows($result_set) > 0){

  for ($i = 0; $i < mysql_num_rows($result_set); $i++)
  {
     $ename = mysql_result($result_set, $i, "Event_Name");
     $date = mysql_result($result_set, $i, "Event_Date");
     $vname = mysql_result($result_set, $i, "Venue_Name");
     $description = mysql_result($result_set, $i, "Event_Description");
     $cost = mysql_result($result_set, $i, "Event_Cost");
               $img = mysql_result($result_set, $i, "id");

        echo' <div style="margin:9 15 0 20px "><a href="#"><img src="images.php?id=$img" align="left" style="margin-right:15px " alt="" border="0">[/url]<span style="color: #00BCD4"><br style="line-height:9px">
                                             <span class="style4">'.$ename.' at '.$vname.'</span></span>

                              '.$description.'... 
<span class="style8">VIP Exclusive Cost of: '.$cost.'</span>

 <span class="style6" style="float:right; margin: 0 20px 0 0"><a href="#"><strong>read more</strong>[/url]</span></div>

    <br style="line-height:11px"><img src="images/1_line.gif" style="margin-left:55px " alt="" border="0" width="425">
<br style="line-height:10px">';

  }
  } else {
     echo 'Sorry, no results were found...';
  }
?>

 

Good luck!

[suess0r]

dbillings, thanks so much... clear and well explained explanation. Everything's working great, thanks again!

[dbillings]

Click solved on your thread.