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Programming  »   How to edit this script to make images and urls clickable

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  1.  
    • CommentAuthorPiKa88
    • CommentTimeOct 23rd 2009 edited
     

    The site is

    Code:
    http://hosttops.com



    Lets say i go to the main page "most recent reviews" and i just scroll down to the first listing. It will have an image next to it and if i click on that image it will open a new window and display the image? How do i change this so that when i click on the image it will take me to the page that was being reviewed, the url which links to another website.

    and also, if i look at that same page, "most recent reviews", I will see a listing like this:

    Review Date: Jan 21, 2009
    URL:

    Code:
    http://affiliatepage.com


    Plan Bandwidth: 5GB
    Plan Price: $6
    Plan Space: 5GB

    how do i change that url so that it is clickable as a link?
    like this:
     

     

    <a href="http://an-affiliate-link-to-the-web-host.com">http://The-Web-Host.com</a>

     



    Please help, Thanks!

    • CommentAuthorPiKa88
    • CommentTimeOct 23rd 2009 edited
     

    Look in the scripts source code, it should be grabbing data from MySQL and then echo'ing it to the page somehow. The line would prolly look something like this:


    while($row = mysql_fetch_assoc)
    {
         echo $row['image'];
    }



    Just change the output to output the link to the website inside the <a> tag instead of the image.

    I really can't explain it any better, since I don't have the source code. Good luck 

    • CommentAuthorKameo
    • CommentTimeOct 23rd 2009
     
    Yea, i searched the source code for something like that... hmm i will check the mysql to figure out if the tag is something weird and not something simple like "image"
  4.  

    Ok the images are located under "item_media" in the mysql database and to me it seems like this is the code that displays the image onto the main page 

    $sql2="select * from ".TABLE_PREFIX."item_media where i_id='$this->pid'$And order by file_id asc"; 
    //echo $sql2; 
    $valsql = $this->_db->query($sql2); 
    $img = ''; 
    if($this->_db->num_rows($valsql) > 0){ 
    while($rows=$this->_db->fetch_array($valsql)){ 
    if (!empty($rows['isFile'])){ 
    $file = explode(".",$rows['file_path']); 
    $this->finame = $file[0]; 
    $this->fullname = $rows['file_path']; 
    $this->filetype = $rows['itype']; 
    } else { 
    if ($f_image==""){ 
    $f_image=$rows['file_id']; 
    } 
    //get the detail image name 
    $this->file_id=$rows['file_id']; 
    $this->LoadPhoto(); 
    $img.=$rows['file_id']."|".$this->dtl_image."++";//."|".$rows[image_name]."++"; 
    } 
    } 
    $this->_db->free_result($valsql); 
    } 

    if ($img!=""){ 
    $img=substr($img,0,-2); 
    } 
    $this->img_str=$img; 
    if ($f_image!=""){ 
    $this->file_id=""; 
    $this->file_id=$f_image; 
    $this->LoadPhoto(); 
    }

    However i am not sure of that, and im not sure what to change, I will mess with it and see if i can get it to work. 

    Thanks for the help, and let me know if you can figure it out

     

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