RETRIEVING Image 4RM DB Using Image ID

vexer007

Hi PLEASE HELP ME,

I am trying to populate a table with images using database. So I have these columns in my DB Tilte, Price, Image.

I already can display the price and the tiltle in a Table row, now I also want the Images display next to the price and title.

GOOD NEWS!!! I can now display an image from the database because I put this in the DB colomn
images/tony_blair.jpg and then use php to display the column value.

PROBLEM??? How can I now display multiple rows obviously the images differ, do i have to go to the columns and keep typing

images/whatever.jpg
images/something.jpg

Or I was told I can use IDs How can I Do THAT? Help Please

Thank you for Your Time! Am new to PHP Using Acess DB

bastien

put it in a loop...the sample below is to pull imges from a db, which you are not doing, so you would just sub the correct element into the image tag

<?php

// show_desc.php
require("conn.php");

  // you may have to modify login information for your database server

  $query = "select description, id from binary_data ";
  $result = connect($query);

  while ($rows = MYSQL_FETCH_ARRAY($result)) 
  { 
  echo $rows['description']; 
  echo "<br><br>";

  echo "<img src=\"show_image.php?id=".$rows['id']."\">\n";


  }
?>

vexer007

Hi bastien, Thank you for That Excellent tip! I am very greatful.

few clarifications please

in the echo tag below, dont I have to tell where the images folder is i.e echo "<img src=images\show_image.php?id=".$rows['id']."\">\n";

or can you just help a little bit more to clarrify what the show_image.php does? is it a pre defined function?

echo "<img src=\"show_image.php?id=".$rows['id']."\">\n";

I understand your given code to be selecting two columns from my binary_data table and if row is found, display the rows description column data and display the image from ? where the image name = the column ID am I right??? please tell me I am? Where would my images be kept?

Thanks a million for this help! I appreciate it

bastien

show_images.php is a page that takes the image stored in the db and sends it to the browser as part of another page...you have to do this with images stored in a db because the page disposition headers are diff for regular html and images.
[id] is the autonumber record id of the record to be displayed. I tend to use and autonumber primary key. Passing this key id to the show_images.php tells that page which row in the db the image resides in

vexer007

Stored Image in db? I was thinking if it was possible to store the images in an images folder and giving them names in relative to the autonumber ID field so if I say Display image from /images where image name = Auto number ID? no? Sorry am a bit comfuse please, If you say "images stored in db" does that mean in my db I should have a column for images and simply populate it with /images/imagename.jpg ? Finally in the Image retrieving page what will that contain,
<connection to db>
sql=" select autonumber, image_id from image_table where ?
<img src="something.jpg">

Am a bit lost? Your first example was excellent, please help me a little bit more by clarifying it slightly as am quite new to php.

Thanks a Million for your valuable time

bastien

my bad...i assumed from the title of your post and its content that you wanted the image stored in the db too.

Since you don't, you only need to change a line


<?php

  echo "<img src=\"path/to/image/.$rows['image_name']."\">\n";


  }
?>

vexer007

So How do I ensure that the right image is displayed next to the right price? i.e associate an image to an autonumber ID so as to say display image from image folder where image_name = db's product_Table's autonumber? Or is that not neccessary? I often see image id in codes, how does that come in?

Thank you, the code is much clearer by the way now as well.
You've been very helpful indeed.

bastien

The select * sql code selects all the data together, those rows won't shuffle arround during the display processing. So images and the associated other data (price, desc etc) will all stay together when that row is handled in the code


echo "<table>";
while ($rows = MYSQL_FETCH_ARRAY($result)) 
  {
 // put it in a table 
  echo "<tr><td>".$rows['description']; 

  echo "<td><img src=\"show_image.php?id=".$rows['id']."\">\n";

  echo "</td>";
  }

vexer007

I applaude you! you are good, Can't wait to be a better programmer like you. You've been more than helpful. Much appreciated.

Thanks again for your time...