Display one of three images in a table cell based on a field in my database

[maxwell_]

I am trying to display one of three images in a table cell based on a field in my database.

I have created the following function

 function chooseImage($imagetype)
    {
		 global $connection;
		$sql = "select * from uploads";
$result = $connection->query($sql) or die(mysqli_error($connection));
$numRows = $result->num_rows;

	 while($row = $result->fetch_assoc()) { // opening while loop bracket tag flag 2

 $TypeOfFile = $row['document_type'];

		if($TypeOfFile == 'PDF'){

		 $imagetype = "Adobe_Acrobat.png";				

			}elseif($TypeOfFile == 'JPG'){

			$imagetype = "jpg.png";

				}elseif($TypeOfFile == 'WORD') {

			 $imagetype = "iconBG4.gif";	
					} 
	 }

	return $imagetype;
}

which I call as follows

$file_type = chooseImage($imagetype);

<td>

<img src="images/<?php echo $file_type; ?>" alt="">    

<td>

However the same image is displaying even though the database field contains WORD of JPG.
I suspect it is being caused by the way I am returning the variable within the function.

Best regards Maxwell

[Weedpacket]

So.... how is the function supposed to choose which filename to return? You're just telling it to loop through all of the rows in the uploads table and you're keeping the last one.

[maxwell_]

Can you please give a bit more advice on the proper way to go about this.

Best regards Maxwell

[dalecosp]

You need a separate icon for each image, correct? It seems to me it'd make more sense to pass an ID to the function:

function chooseIcon($image_id) 
{ 
    global $connection; 
    $sql = "select document_type from uploads where id = $image_id"; 
    $result = $connection->query($sql) or die(mysqli_error($connection)); 
    $numRows = $result->num_rows; 

while($row = $result->fetch_assoc()) // opening while loop bracket tag flag 2 
{
    $TypeOfFile = $row['document_type']; 

    if($TypeOfFile == 'PDF')
    { 
        return "Adobe_Acrobat.png";                 
    }
    elseif($TypeOfFile == 'JPG')
    { 
        return "jpg.png"; 
    }
    elseif($TypeOfFile == 'WORD')
    {                                                 
        return "iconBG4.gif";     
    }  
} 
return ''; 
}

[maxwell_]

Thanks for the advice.

I managed to do it using a variable defined from an earlier query in my script which contained the document_type database field

$document_type = $row['document_type'];

$query = "SELECT * FROM document_type WHERE type_id = {$document_type} ";
							$select_type = mysqli_query($connection, $query);

				if (mysqli_num_rows($select_type) > 0) {

						while($row = mysqli_fetch_assoc($select_type)){

						$type_id = $row['type_id'];
						$file_type = $row['file_type'];	

						if($file_type == 'PDF') 
						{  
							$file_type = 'Adobe_Acrobat.png';                 
						} 
						if($file_type == 'JPG') 
						{  
							$file_type = 'jpg.png';                 
						}
						elseif($file_type == 'WORD') 
						{  
							$file_type = 'word.png';                 
						} 

<img class="imgBottom" src="images/<?php echo $file_type; ?>"> 	

 }  

						} 

[NogDog]

Just a note, you don't really need to switch between [noparse]

 and [code=html][/noparse] bbcode tags here as your code switches between them. If it's a PHP file, probably best to just use the PHP tags. :)