Returning a value from a function call - variable scope issues!

[conjurer]

I have a script where I am calling a function to query a database and pull the filename that is stored there.

In the code below it is the last block that is giving me problems.

The echo command sends the correct file name to the screen. But when I leave the function and try to use the variable $img_name I get an error message that this is an undefined variable:

Encountered error: 8 in /home/omgma/public_html/member_community/profile/profile_updt_form.php, line 181: Undefined variable: img_name

function get_img_filename($username)
	{
		$conn = db_connect();

	$result = $conn->query("select user_img_filename from users where username = '$username'");

		  $row = $result->fetch_object();
		  $img_name = $row->user_img_filename;
		  echo "This is image name: $img_name";
		  return $img_name;


}

The script where I call the function is as follows:

 <?php 
            //show current photo if exists
            //This will be a query of the database to see if there is an existing image name stored in db 

      	$imgs_path = '/home/omgma/public_html/member_community/profile/profile_imgs/' ;

		//if there is then show the existing photo
        //check for the $newfilename variable

            if (!empty($newfilename))
                {
                //$newfilename is not empty so use name in processing
                $imgs_path .= $newfilename  ; 
                }
            else
                {//$newfilename is empty so find the filename in db

				    $username = $_SESSION['valid_user']; 
					get_img_filename($username);
	    		 }

              	//check to see if a user photo exists
				$imgs_path .= $img_name  ;
                if (file_exists($imgs_path) )
                {
                print "<div><img src='profile_imgs/".$img_name ."' alt='".$username."'  /></div>";

				print "<p>Path and nameVariables: </p>";
                var_dump ($imgs_path, $img_name);


		 		}

                else
                {
                print "<div><img src='profile_imgs/no_photo.jpg' alt='No Image Available' /></div>";
        		/* 
				print "<p>No image could be found - path and name variables:</p>";
                var_dump ($imgs_path, $newfilename);
				*/
				}	
		?>

So how do I make the img_name variable that is set to the query results in the function available to the script that calls the function?

[laserlight]

So how do I make the img_name variable that is set to the query results in the function available to the script that calls the function?
You assign the return value of the function call, e.g.,

$img_name = get_img_filename($username);

However, I note that you only call the function in the else block.

[conjurer]

You are correct on the else block - here is the whole function below.

I do some error checking first and if it passes through the error trapping blocks then a filename was found and I want to return that.

Would I still be able to do that by assigning the variable name to the function?

//---------------------------------------------------------------------------
// function to get user image filename
//---------------------------------------------------------------------------

function get_img_filename($username)
	{
		$conn = db_connect();

	$result = $conn->query("select user_img_filename from users where username = '$username'");
	if (!$result)
		{//could not execute query
		  throw new Exception('<h2>Could not execute query to find the member profile image!</h2><p>Error: '
								.mysqli_errno($conn)
								.mysqli_error($conn)
								.'</p>' );
		}
	else if ($result->num_rows==0)
		{// file not found
		  throw new Exception('<h2>Could not find image name in database! </h2><p>Error: '
								.mysqli_errno($conn)
								.mysqli_error($conn)
								.'</p>' );
		}


// <<<This is where I want to grab the name and return the value>>>>
		else
			{

		  $row = $result->fetch_object();
		  $img_name = $row->user_img_filename;
		  echo "This is image name: $img_name";
		  return $img_name;
		}


}//Function end

[laserlight]

Well, you need to improve your indentation: I'm getting confused as to what exactly you are trying to do in your code.

The point is, you need to make use of the return value, not just call the function.

[conjurer]

Thank you for trying to help. Not sure what you think is wrong with my indentations.l I will agree it is a complicated block of code but indentation is fine and it probably isn't helpful to a newbie to tell them to improve their indentation.

I tried assigning a variable to the function and that is working if the field has a value in the database. When blank it is messing up. I will do a work around on that part. Thanks again for help.

[laserlight]

conjurer wrote:

Not sure what you think is wrong with my indentations.l I will agree it is a complicated block of code but indentation is fine and it probably isn't helpful to a newbie to tell them to improve their indentation.

It is not a very complicated block of code, but the poor indentation makes it into a mess that looks complicated 😉

This is how I might format the code for the function:

function get_img_filename($username)
{
    $conn = db_connect();

$result = $conn->query("select user_img_filename from users where username = '$username'");
if (!$result)
{
    //could not execute query
    throw new Exception('<h2>Could not execute query to find the member profile image!</h2><p>Error: '
        . mysqli_errno($conn)
        . mysqli_error($conn)
        . '</p>' );
}
else if ($result->num_rows == 0)
{
    // file not found
    throw new Exception('<h2>Could not find image name in database! </h2><p>Error: '
        . mysqli_errno($conn)
        . mysqli_error($conn)
        . '</p>' );
}
else
{
    // <<<This is where I want to grab the name and return the value
    $row = $result->fetch_object();
    $img_name = $row->user_img_filename;
    echo "This is image name: $img_name";
    return $img_name;
}
}

=conjurer wrote:

I tried assigning a variable to the function and that is working if the field has a value in the database. When blank it is messing up.

What exactly do you mean by "assigning a variable to the function"?