Passing A Variable outside a Function?

[spires]

Hi Guys.

 

I have created a function that allows me to upload files to my server.

However, I need to pass the file name to outside of the function, how is this done.

 

Thanks

 

function details_image1_upload($Iname){
if ($_FILES['image']['name']==''){

}else{

     $filedir = 'images/details/1/'; // the directory for the original image

     $maxfile = '1000000'; // max file size
     $mode = octdec('0666'); // octdec -- Octal to decimal.
 						// The mode parameter consists of three octal number components specifying access restrictions for the owner,
						// the user group in which the owner is in, and to everybody else in this order. e.g (666)
      
     $userfile_name = $_FILES['image']['name']; 
     $userfile_tmp = $_FILES['image']['tmp_name']; 
     $userfile_size = $_FILES['image']['size']; 
     $userfile_type = $_FILES['image']['type'];
    $userfile_error = $_FILES['image']['error'];

	 $Itype = 'jpg';
		$strName =  str_replace(' ', '_', $Iname);
	 $Dimg1 = $strName.".".$Itype;

 	// $prod_img = the image folder and the image and name
         $prod_img = $filedir.$Dimg1; 


	 //move_uploaded_file -- Moves an uploaded file to a new location.
	 //move the uploaded file to the image folder with a tempory name.
         if(move_uploaded_file($userfile_tmp, $prod_img)){
        $in =  "Thank You, Your File Has Been Uploaded";
         }
         else{
        $out =  "Sorry, Your File Failed To Upload";
         }
	 // chmod -- Changes file mode, 
	 // set the user interface for the image
         chmod ($prod_img, $mode); 
     
				global $Dimg1;
				} 
}


echo $Dimg1;

[premiso]

You can use $_SESSION to do that, or set the $filename as a global.

 

http://us3.php.net/manual/en/language.variables.scope.php  for more details on the global part.

[.josh]

I do not recommend making variables global inside an array.  Defeats the purpose of having scope. 

 

It depends on what you mean by "passing the file name outside the function".

 

Are you calling another function inside the function? Just pass it as an argument in the function call.  Are you wanting to have this file name after the function is called, and its done it's business? Just return the file name.

 

function blah() {
  $filename = "somefile.php";
  return $filename;
}

$filename = blah();

echo $filename; // output: somefile.php

[spires]

Hi Guys,

 

Thanks for your help.

 

This is what I did

// Call Main Image Upload
details_image1_upload($_POST['Iname1']);
// Call 2nd Image Upload
details_image2_upload($_POST['Iname2'])	;


$Dimg1 = details_image1_upload($_POST['Iname1']);
$Dimg2 = details_image2_upload($_POST['Iname2']);

 

However, how do I get three variable that were created inside a function, to be used in the rest of my script?

 

Thanks

[Maq]

Return an array instead.

 

function blah() {
  $filename[] = "somefile1.php";
  $filename[] = "somefile2.php";
  $filename[] = "somefile3.php";

  return $filename;
}

$filename = blah();

foreach($filename as $key)
   echo $key . "
";