$_FILES for file uploads in PHP - Mini Project: Build & Apply

Handling File Uploads with $_FILES in PHP

📖 Scenario: You are building a simple web page where users can upload a profile picture. You want to learn how PHP handles uploaded files using the $_FILES superglobal.

🎯 Goal: Create a PHP script that accepts a file upload from an HTML form, accesses the uploaded file information using $_FILES, and displays the original file name and file size.

📋 What You'll Learn

Create an HTML form with a file input named profile_pic and method POST with enctype="multipart/form-data".

Create a PHP script that accesses the uploaded file data using $_FILES['profile_pic'].

Extract the original file name and file size from $_FILES['profile_pic'].

Display the file name and size in bytes on the page.

💡 Why This Matters

🌍 Real World

File uploads are common on websites for profile pictures, documents, or other media. Understanding <code>$_FILES</code> helps you handle these uploads safely and effectively.

💼 Career

Many web developer jobs require handling file uploads securely. Knowing how to use <code>$_FILES</code> is essential for backend PHP development.

1

Create the HTML form for file upload

Write the HTML code to create a form with method="POST" and enctype="multipart/form-data". Inside the form, add a file input with the name profile_pic and a submit button labeled Upload.

PHP

Hint

Remember to set enctype="multipart/form-data" on the form to allow file uploads.

2

Check if a file was uploaded

In PHP, create an if statement that checks if $_FILES['profile_pic']['error'] is equal to UPLOAD_ERR_OK to confirm a file was uploaded without errors.

PHP

    
    Upload

Hint

Use isset($_FILES['profile_pic']) to check if the file input exists and compare the error code to UPLOAD_ERR_OK.

3

Extract the uploaded file's name and size

Inside the if block, create two variables: $fileName to store $_FILES['profile_pic']['name'] and $fileSize to store $_FILES['profile_pic']['size'].

PHP

    
    Upload

Hint

Use the name and size keys inside $_FILES['profile_pic'] to get the original file name and size in bytes.

4

Display the uploaded file's name and size

Add echo statements inside the if block to print the file name and file size in bytes in this format: File name: [file name] and File size: [file size] bytes.

PHP

    
    Upload

Hint

Use echo to print the file name and size with a line break between them.

Practice

1. What does the $_FILES superglobal in PHP store?

easy

A. Database connection details

B. User session data

C. Information about files uploaded via an HTML form

D. Server environment variables

Solution

1. Step 1: Understand the purpose of $_FILES

   $_FILES is a special PHP variable that holds details about files uploaded through a form.

2. Step 2: Compare with other options

   User session data is stored in $_SESSION, database details are not stored here, and server variables are in $_SERVER.

3. Final Answer:

   Information about files uploaded via an HTML form -> Option C

4. Quick Check:

   $_FILES = Uploaded file info [OK]

Hint: Remember: $_FILES always relates to uploaded files [OK]

Common Mistakes:

• Confusing $_FILES with $_SESSION
• Thinking $_FILES holds server info
• Assuming $_FILES stores form text inputs

2. Which of the following is the correct way to access the name of an uploaded file with input name userfile?

easy

A. $_FILES['userfile']['filename']

B. $_FILES['userfile']['file']

C. $_FILES['userfile']['file_name']

D. $_FILES['userfile']['name']

Solution

1. Step 1: Recall the structure of $_FILES

   Each uploaded file's info is stored in an array with keys like 'name', 'type', 'tmp_name', 'error', and 'size'.

2. Step 2: Identify the correct key for the original filename

   The key for the original filename is 'name', not 'filename' or 'file_name'.

3. Final Answer:

   $_FILES['userfile']['name'] -> Option D

4. Quick Check:

   Filename key = 'name' [OK]

Hint: Use 'name' key to get original filename from $_FILES [OK]

Common Mistakes:

• Using 'filename' or 'file_name' instead of 'name'
• Confusing keys with other $_FILES elements
• Trying to access $_POST for file names

3. What will the following PHP code output if a file named 'photo.jpg' is uploaded successfully?

if ($_FILES['pic']['error'] === 0) {
    echo $_FILES['pic']['name'];
} else {
    echo 'Upload failed';
}

medium

A. photo.jpg

B. Upload failed

C. 0

D. Undefined index error

Solution

1. Step 1: Check the error code for the uploaded file

   An error code of 0 means the upload was successful with no errors.

2. Step 2: Output the original filename

   Since error is 0, the code echoes the original filename stored in $_FILES['pic']['name'], which is 'photo.jpg'.

3. Final Answer:

   photo.jpg -> Option A

4. Quick Check:

   Error 0 means success, so filename prints [OK]

Hint: Error 0 means success; print 'name' to get filename [OK]

Common Mistakes:

• Assuming error 0 means failure
• Printing error code instead of filename
• Not checking if 'pic' index exists

4. Identify the error in this PHP snippet handling file upload:

if ($_FILES['upload']['error'] == 0) {
    move_uploaded_file($_FILES['upload']['name'], 'uploads/' . $_FILES['upload']['name']);
    echo 'File saved';
} else {
    echo 'Error uploading';
}

medium

A. Not checking file size before moving

B. Using 'name' instead of 'tmp_name' in move_uploaded_file

C. Checking error with == instead of ===

D. Missing semicolon after echo statement

Solution

1. Step 1: Understand move_uploaded_file parameters

   The first parameter must be the temporary file path stored in tmp_name, not the original filename.

2. Step 2: Identify the incorrect usage

   The code incorrectly uses $_FILES['upload']['name'] as the source file, which is wrong and causes failure.

3. Final Answer:

   Using 'name' instead of 'tmp_name' in move_uploaded_file -> Option B

4. Quick Check:

   move_uploaded_file needs tmp_name as source [OK]

Hint: Use 'tmp_name' as source file in move_uploaded_file [OK]

Common Mistakes:

• Using original filename instead of temp file path
• Ignoring error codes before moving files
• Not validating file size or type

5. You want to save an uploaded file only if it is less than 2MB and has no errors. Which code snippet correctly checks these conditions before moving the file?

if (/* condition here */) {
    move_uploaded_file($_FILES['file']['tmp_name'], 'uploads/' . $_FILES['file']['name']);
    echo 'Upload successful';
} else {
    echo 'Upload failed';
}

hard

A. if ($_FILES['file']['error'] === 0 && $_FILES['file']['size'] < 2097152)

B. if ($_FILES['file']['error'] == 1 || $_FILES['file']['size'] > 2097152)

C. if ($_FILES['file']['error'] !== 0 && $_FILES['file']['size'] <= 2097152)

D. if ($_FILES['file']['error'] === 0 || $_FILES['file']['size'] > 2097152)

Solution

1. Step 1: Check for no upload errors

   The error code must be exactly 0 to confirm no errors during upload.

2. Step 2: Check file size limit

   The file size must be less than 2MB (2 * 1024 * 1024 = 2097152 bytes).

3. Step 3: Combine conditions correctly

   Both conditions must be true, so use logical AND (&&) with strict equality and less-than size check.

4. Final Answer:

   if ($_FILES['file']['error'] === 0 && $_FILES['file']['size'] < 2097152) -> Option A

5. Quick Check:

   Error 0 and size < 2MB = correct check [OK]

Hint: Use error === 0 and size < 2MB with && [OK]

Common Mistakes:

• Using || instead of && for conditions
• Checking error != 0 instead of === 0
• Confusing size units or limits