manage image display

robin01

Hello,

I'm in the process of creating a classified script and working on adding images to the buy and sell part of the script.

I added a few test images in a table and they all seem kind crushed or distorted. no image management script is applied to these images and i'm assuming i need to do this to control the the way image is displayed.

How do i control the size/shape of the images being uploaded.
how do i control how the image is displayed in terms of hight and width without the image getting distorted.

could use some ideas and maybe php scripts that manage the upload and the display..

Thanks
robin

Roger_Ramjet

Storing images in tables is a bad idea for many reasons. Save th images files in a directory and just store the name in the db. When you output the html the image file url is what goes into the html tag anyway. Saves a whole lot of processing and avoids this problem.

robin01

Roger Ramjet wrote:
Storing images in tables is a bad idea for many reasons. Save th images files in a directory and just store the name in the db. When you output the html the image file url is what goes into the html tag anyway. Saves a whole lot of processing and avoids this problem.

Sorry about the misscommunication..

the images are in a folder called images. i should have been more clear. i'm pulling in the image ina html table to display

Roger_Ramjet

And the html table is squashing the image. OK, how are you embedding the image into the table - show us the code.

robin01

Roger Ramjet wrote:
And the html table is squashing the image. OK, how are you embedding the image into the table - show us the code.

Basic the name of the image is stored in the database and this script pretty much pulls in the name and finds the image in the /image folder :

random_ad_display($feature_result);
function random_ad_display($feature_result)
{  #$imageurl = preg_replace("/[^\/]*\.php/","", $_SERVER['REQUEST_URI']). "images/";
   echo "<table width='500' align='center' border='0' cellspacing='0' cellpadding='0'>";
    echo "<tr>";
    $i = 0;
    while($row1 = mysql_fetch_array($feature_result))
    {   if ($i == 2)
        {  echo "</tr>";
           echo "<tr>";
           $image1 = $row1["cf_images"];
           echo "<td width='250' height='120'> <div align='center'> <p align='center'><IMG SRC='images/" . $image1 . "'width='150' height='120' /></p> <p>description and ad number</p></td>";
           $i = 0;
        }
        else
        {  $image2 = $row1["cf_images"];
           echo "<td width='250' height='120'> <div align='center'> <p align='center'><IMG SRC='images/". $image2 . "' width='150' height='120' /> </p> <p> test description 2 testing one two three four five </p></td>"; }
        $i++;
    }    echo "</tr>";
    echo "</table>";
}

Roger_Ramjet

So, is that the true size of the image?

robin01

Roger Ramjet wrote:
So, is that the true size of the image?

hello,

I'm assuming your asking about this line -->> "<td width='250' height='120'> <div align='center'> <p align='center'><IMG SRC='images ...

no, that the dimensions of the table that i need the image to fit in. i have no script to control the image being displayed.

Roger_Ramjet

No I meant this: SRC='images/". $image2 . "' width='150' height='120'
If the ratio of the original image is not the same as that ratio (150:120) then the image will be distorted in the display.