display Alternate image

[proctk]

This one has been giving me a hard time for a long while and I really need to help to get to the bottom of it.

The goal is to display the users photo if one is found and if no photo is found then show the default "blankImage" if no image was found. The below displays both the blank image and the users image if an image is found. any help with this is greatly appreciated

 //Gets buddies photo from thge table images if one exsists
 $query_get_photo = ("SELECT * FROM `image_files` WHERE `user_id` = '$buddy_id' AND `album` = 'Member'");
 $get_photo = mysql_query($query_get_photo)or die("SQL Error: $query_get_photo <br>" . mysql_error());

 while($photo = mysql_fetch_assoc($get_photo)){
 $photoCount = mysql_num_rows($get_photo);
echo $photoCount; 
 ?>

            <?php
	if($photoCount > 0){
 	$image = "user_images/".$photo['image_name'];
	$tempimg = getImageSize($image);
	$size = $tempimg[3];
	$size = explode("\"",$size);
	$width = $size[1];
	$height = $size[3];

	$newwidth=75;
	$newheight=($height/$width)*75;

	?>
            <a href="profile.php?id=<?php echo $buddy_id; ?>"><img src="user_images/<?php echo $photo['image_name'];?>"width="<?php echo $newwidth; ?>" height="<?php echo $newheight?>" alt="<?php echo $photo['image_name'];?>" /></a>
            <?php }else{ ?>
            <a href="profile.php?id=<?php echo $buddy_id; ?>"><img src="user_images/blankAlbum.jpg" width="75" height="75" alt="blankAlbum.jpg" /></a><a href="profile.php?id=<?php echo $buddy_id; ?>"></a><?php } }?>

[proctk]

I just did another post similar to this one, sorry about that, I did not realize there was a delay.

[proctk]

I just figured out a huge problem. not every person will be found in the table image_files. I need to figure out how to change my if statememt so that if the buddy id was not found in the table then display the blank image.

I have edited my code to this which is not working because the $photo['user_id'] != $buddy_id) statement does not work as $photo['user_id'] != $buddy_id) has no value.

the bottom line is if the nothing is found in the query then I want the blank image to display

any help is excellent

 if ($photo['user_id'] == $buddy_id){
		$imageName = $photo['image_name'];

	$image = "user_images/".$imageName;
	$tempimg = getImageSize($image);
	$size = $tempimg[3];
	$size = explode("\"",$size);
	$width = $size[1];
	$height = $size[3];

	$newwidth=75;
	$newheight=($height/$width)*75;
	}

	elseif($photo['user_id'] != $buddy_id){
	$imageName = 'blankAlbum.jpg';
	$newwidth=75;
	$newheight=100;
	}

[proctk]

This will display the image but gives an error message for the spots where the blank image is supposed to display

ERROR
Warning: Invalid argument supplied for foreach() in C:\xampp\htdocs\familyclick\buddyList.php on line 120
any yet another approach to this

My thought is the error is because there is not the same number of buddy_ids as $ids. I have no clue how to fix this

any help is great thank you

<div id="Centercontent">

      <p>
        <?php while($buddy = mysql_fetch_assoc($get_buddies)){ // gets users buddies
 $buddy_id = $buddy['buddy_id'];


  // Get buddies information for table user
 $query_get_buddy_info = ("Select * From `users` WHERE `user_id` = '$buddy_id'");
 $get_buddy_info = mysql_query($query_get_buddy_info )or die("SQL Error: $query_get_buddy_info <br>" . mysql_error());

 while($buddy_info = mysql_fetch_assoc($get_buddy_info)){?>

             <div class="results">

     	<ul class="resultLinks">
    		<li><a href="profile.php?id=<?php echo $buddy_id; ?>">View Profile</a></li>
            <li><a href="#">Send Message</a></li>
             <?php $query_checkfor_buddy = ("SELECT * FROM buddylink WHERE owner_id = '$buddy_id'"); 
			$checkfor_buddy = mysql_query($query_checkfor_buddy)or die("SQL Error: $query_checkfor_buddy<br>" . mysql_error());
			$buddyCount = mysql_num_rows($checkfor_buddy);

			if($buddyCount >0){
			?>
            <li><a href="buddyList.php?id=<?php echo $buddy_id; ?>">View Buddies</a></li>

            <?php } ?>

            <?php $query_checkfor_albums = ("SELECT * FROM albumnames WHERE user_id = '$buddy_id'"); 
			$checkfor_albums = mysql_query($query_checkfor_albums)or die("SQL Error: $query_checkfor_albums<br>" . mysql_error());
			$albumCount = mysql_num_rows($checkfor_albums);

			if($albumCount >0){
			?>
            <li><a href="photos/buddyAlbums.php?id=<?php echo $buddy_id; ?>">Photo Albums</a></li>
            <?php }?>
            <li><a href="#">Delete Buddy</a></li>
        </ul>
        <div class="resultImg">
                 <div align="center">
            <?php

 $arrbuddy_id = array($buddy_id);

 //Gets buddies photo from thge table images if one exsists
 $query_get_photo = ("SELECT * FROM `image_files` WHERE `user_id` = '$buddy_id' AND `album` = 'Member'");
 $get_photo = mysql_query($query_get_photo)or die("SQL Error: $query_get_photo <br>" . mysql_error());

while ($row = mysql_fetch_array($get_photo, MYSQL_ASSOC)) {
    if(is_null($row['user_id']) || empty($row['user_id']) || $row['user_id'] < 1){
        // do nothing
    }else{
        $ids[] = $row['user_id'];
        $image_file[] = $row[image_name];
    }
} 


  foreach($ids as $value){
  	foreach ($image_file as $file){
	  if ($value == $buddy_id){
		$imageName = $file;

	$image = "user_images/".$imageName;
	$tempimg = getImageSize($image);
	$size = $tempimg[3];
	$size = explode("\"",$size);
	$width = $size[1];
	$height = $size[3];

	$newwidth=75;
	$newheight=($height/$width)*75;

	}

	if($value != $buddy_id){
	$imageName = 'blankAlbum.jpg';
	$newwidth=75;
	$newheight=100;

	}
		}
			}



	?>
            <a href="profile.php?id=<?php echo $buddy_id; ?>"><img src="user_images/<?php echo $imageName ;?>"width="<?php echo $newwidth; ?>" height="<?php echo $newheight?>" alt="<?php echo $imageName;?>" /></a>
            <?php
			?>
          </div>
        </div>
      <div class="resultMiddle">
      <dl class="clearPaddingMargins">
      <dt><span class="resultSubheader">Name:</span></dt> <dd class="resultText"><a class="linkButton" href="profile.php?id=<?php echo $buddy_id; ?>"><?php echo $buddy_info['first_name'].' '.$buddy_info['last_name']; ?></a></dd>
      	<dt><span class="resultSubheader">Location:</span></dt><dd class="resultText"><?php echo $buddy_info['city']. ' '.$buddy_info['province']; ?></dd>

     </dl>
      </div>
      <div class="clear"></div>
      </div>

                <?php 
	  		}
	  			}?>

[mrgrammar]

Will this solve your id issue?

while ($row = mysql_fetch_array($get_photo, MYSQL_ASSOC)) {
    if(is_null($row['user_id']) || $row['user_id'] == "" || $row['user_id'] < 1){
        // do nothing
    }else{
        $ids[] = $row['user_id'];
        $image_file[] = $row[image_name];
    }
}

That's very rough and needs to be refined, but is just meant to communicate the concept. Test and see if $row['user_id'] has data. If not, exclude it from the array. Then you won't get the foreach error.

You can use the same methodology for your blank image.

if(file_exists($image_file[]){
    $imagename = $image_file;
}else{
    $imagename = blankimagename.gif
}

For more information about checking for the existence of the file, see http://us3.php.net/manual/en/function.file-exists.php

Hope that helps.

[proctk]

I'm thinking that I did not explain myself very well

The person could have a photo in the table file_images. note its just the file name saved in the mysql table and not the photo the photo is saved in a folder called user_images.

Its okay if the user does not have a photo for the folks that don't have photos I want the file "blank.jpg". to display.'

The last snippet of code I posted will show the photo is one is found for the user but if there was no photo found its giving an error message are because of the foreach statement.

I think the problem is there are more buddy then photo[user_ids] and the loop has nothing to compare to.

hope this helps explain it better and thnk you for taking time to reply

cheers
kp

Note I edited the code I posted above to show all the code this might help explain more

I tried to code you posted above and I get the same foreach error message

[proctk]

I continue to search on this one.

I came across array_diff and array_intersect. I have been trying to use these two functions but its the array ids keeps returning an error message where there is not match for intersect.

so I tried changing $ids to $arrids = ids

this kinda helped, it cleared all the error message but its adding a second array to the matches.

while ($row = mysql_fetch_array($get_photo, MYSQL_ASSOC)) {
    if(is_null($row['user_id']) || $row['user_id'] == "" || $row['user_id'] < 1){
        // do nothing
    }else{
        $ids[] = $row['user_id'];
        $image_file[] = $row[image_name];
    }
} 

$arrBuddy_id = array($buddy_id);
$arrids = array($ids);

$notfoundId = array_diff($arrBuddy_id,$arrids);

$foundIds  = array_intersect($ids, $arrBuddy_id);

print_r($foundIds);

foreach($foundIds as $value){
if($value == $buddy_id){
		$imageName = 'blankAlbum.jpg';
		$newwidth=75;
		$newheight=100;
}
}

[proctk]

Alright Its been days and I'm still banking my head on this one.

The code should work but there is only one buddy_id showing when I echo $buddy_id within the while loop if I echo it outside the while loop all buddy Ideas show.

I tried assigning the buddy Is to an array to bring them inside the while loop but its only returning the buddy ids that are found in the query

help please I'm getting desperate

$arrBuddy_id = array($buddy_id);

 //Gets buddies photo from thge table images if one exsists
 $query_get_photo = ("SELECT * FROM `image_files` WHERE `user_id` = '$buddy_id' AND `album` = 'Member'");
 $get_photo = mysql_query($query_get_photo)or die("SQL Error: $query_get_photo <br>" . mysql_error());

while ($row = mysql_fetch_array($get_photo, MYSQL_ASSOC)) {
print_r($arrBuddy_id);
$countIds = mysql_num_rows($get_photo);
	if($buddy_id == $row['user_id']){

	$image = "user_images/".$row['image_name'];
	$tempimg = getImageSize($image);
	$size = $tempimg[3];
	$size = explode("\"",$size);
	$width = $size[1];
	$height = $size[3];

	$newwidth=75;
	$newheight=($height/$width)*75;

	?>
            <a href="profile.php?id=<?php echo $buddy_id; ?>"><img src="user_images/<?php echo $row['image_name'] ;?>"width="<?php echo $newwidth; ?>" height="<?php echo $newheight?>" alt="<?php echo $row['image_name'];?>" /></a>
    <?php }} ?>

[proctk]

I was making this way to complicated I solved the issue