ericp Posted August 12, 2013 Share Posted August 12, 2013 Hi there, Is there anyone having the same issue as mine or not that my uploaded file is not renamed in the database though I have tried several uploads? I do not make any changes to the sample codes at all. // Create a temporary file name: $temp = './uploads/' . md5($_FILES['image']['name']); . . . // rename image $id = mysqli_stmt_insert_id($stmt); // Get the print ID rename ($temp, "./uploads/$id"); Your help will be appreciated. Regards, Ericp Link to comment Share on other sites More sharing options...
HartleySan Posted August 12, 2013 Share Posted August 12, 2013 What is $id equal to? Link to comment Share on other sites More sharing options...
Larry Posted August 13, 2013 Share Posted August 13, 2013 It's not renamed in the database or renamed on the file system? The rename() function only impacts the file system. Link to comment Share on other sites More sharing options...
ericp Posted August 13, 2013 Author Share Posted August 13, 2013 It's not renamed in the database or renamed on the file system? The rename() function only impacts the file system. OK. I will take a look at them when the files—the images—are served by the public side of the site, as the book says 'you’ll soon see'. Just a bit curious... as I thought that it's renamed in the database. Link to comment Share on other sites More sharing options...
ericp Posted August 13, 2013 Author Share Posted August 13, 2013 What is $id equal to? To my understanding, it is the variable that carries the value to be retrieved so that the associated image can be renamed later. Am I correct, Larry? Link to comment Share on other sites More sharing options...
ericp Posted August 13, 2013 Author Share Posted August 13, 2013 Hi all, One more thing about this script is that I tried to improve this script by adding the validation code to allow some certain file size and types as per the script 11.2 in the chapter 11 suggests- before moving them to the permanent destination, 'uploads' folder. It works well! My concern is that should I add the else { $temp = NULL; } to the if (is_uploaded_file($_FILES['image']['tmp_name'])) {... or to the if ($_FILES['image']['error'] > 0) {... ? because the code check the size errors before the type ones. Rgds, Link to comment Share on other sites More sharing options...
Larry Posted August 14, 2013 Share Posted August 14, 2013 $id is the print ID, matching the record in the database. The script tries to rename the file to this value. I think HartleySan was trying to ask you what the actual value is, not what the variable is for. Link to comment Share on other sites More sharing options...
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