JSON encoding and decoding in PHP - Mini Project: Build & Apply

JSON encoding and decoding

📖 Scenario: You are working on a simple PHP script that needs to handle user data. You will practice converting PHP arrays to JSON format and back, which is common when sending or receiving data in web applications.

🎯 Goal: Build a PHP script that creates an array of user information, converts it to a JSON string, then decodes the JSON back to a PHP array, and finally prints the decoded data.

📋 What You'll Learn

Create a PHP array with specific user data

Create a variable to hold the JSON encoded string

Decode the JSON string back to a PHP array

Print the decoded array in a readable format

💡 Why This Matters

🌍 Real World

Web applications often send and receive data in JSON format between the server and client. Knowing how to encode and decode JSON in PHP is essential for handling API responses and requests.

💼 Career

Many PHP developer roles require working with JSON data for REST APIs, AJAX calls, and configuration files. This skill is fundamental for backend and full-stack developers.

1

Create the user data array

Create a PHP array called user with these exact key-value pairs: 'name' => 'Alice', 'age' => 30, and 'city' => 'New York'.

PHP

<?php
// Create the user array with name, age, and city
// Your code here

Hint

Use square brackets to create the array and assign the keys and values exactly as shown.

2

Encode the array to JSON

Create a variable called jsonString and set it to the JSON encoded string of the user array using json_encode().

PHP

 'Alice', 'age' => 30, 'city' => 'New York'];
// Encode the user array to JSON and store in jsonString
// Your code here

Hint

Use the json_encode() function and assign the result to $jsonString.

3

Decode the JSON string back to an array

Create a variable called decodedUser and set it to the result of decoding jsonString back to a PHP array using json_decode() with the second parameter set to true.

PHP

 'Alice', 'age' => 30, 'city' => 'New York'];
$jsonString = json_encode($user);
// Decode jsonString back to an array and store in decodedUser
// Your code here

Hint

Use json_decode() with the second argument true to get an associative array.

4

Print the decoded array

Use print_r() to print the decodedUser array so it is readable.

PHP

 'Alice', 'age' => 30, 'city' => 'New York'];
$jsonString = json_encode($user);
$decodedUser = json_decode($jsonString, true);
// Print the decodedUser array
// Your code here

Hint

Use print_r($decodedUser); to display the array in a readable format.

Practice

1. What does the PHP function json_encode() do?

easy

A. Converts PHP data into a JSON formatted string

B. Converts JSON string into PHP data

C. Validates JSON syntax

D. Reads JSON data from a file

Solution

1. Step 1: Understand the purpose of json_encode()

   This function takes PHP variables like arrays or objects and converts them into a JSON string format.

2. Step 2: Differentiate from json_decode()

   json_decode() does the opposite by converting JSON strings back to PHP data.

3. Final Answer:

   Converts PHP data into a JSON formatted string -> Option A

4. Quick Check:

   json_encode() = PHP to JSON string [OK]

Hint: json_encode() makes PHP data into JSON text [OK]

Common Mistakes:

• Confusing json_encode() with json_decode()
• Thinking json_encode() reads JSON from files
• Assuming json_encode() validates JSON syntax

2. Which of the following is the correct syntax to decode a JSON string into a PHP array?

easy

A. $data = json_encode($jsonString, true);

B. $data = json_decode($jsonString, true);

C. $data = json_decode($jsonString);

D. $data = json_encode($jsonString);

Solution

1. Step 1: Identify the function to decode JSON

   The function to convert JSON string to PHP data is json_decode().

2. Step 2: Use the second parameter to get an array

   Passing true as the second argument returns an associative array instead of an object.

3. Final Answer:

   $data = json_decode($jsonString, true); -> Option B

4. Quick Check:

   json_decode() with true returns array [OK]

Hint: Use json_decode($json, true) for PHP arrays [OK]

Common Mistakes:

• Using json_encode() instead of json_decode()
• Omitting the second parameter to get an array
• Passing true to json_encode() which is invalid

3. What will be the output of the following PHP code?

$arr = ['name' => 'Anna', 'age' => 25];
$json = json_encode($arr);
echo $json;

medium

A. {"name":"Anna","age":25}

B. Array

C. name: Anna, age: 25

D. Error: Cannot encode array

Solution

1. Step 1: Understand json_encode output

   The function converts the PHP associative array into a JSON string with keys and values in double quotes.

2. Step 2: Check the exact JSON string format

   The output is a JSON object string: {"name":"Anna","age":25} with escaped quotes when printed.

3. Final Answer:

   {"name":"Anna","age":25} -> Option A

4. Quick Check:

   json_encode() outputs JSON string [OK]

Hint: json_encode() outputs JSON string with quotes [OK]

Common Mistakes:

• Expecting PHP array print instead of JSON string
• Missing quotes around keys and values
• Confusing output with plain text or error

4. Identify the error in this PHP code snippet:

$json = '{"fruit":"apple","count":3}';
$data = json_decode($json, true);
echo $data->fruit;

medium

A. json_decode cannot parse this JSON

B. Incorrect JSON string format

C. Missing second parameter in json_decode

D. Using object syntax on an array after json_decode

Solution

1. Step 1: Check json_decode parameters

   The second parameter true makes json_decode return an associative array.

2. Step 2: Accessing array with object syntax

   Using $data->fruit tries to access an object property, but $data is an array, so this causes an error.

3. Final Answer:

   Using object syntax on an array after json_decode -> Option D

4. Quick Check:

   json_decode(true) returns array, use $data['fruit'] [OK]

Hint: Use $data['key'] if json_decode(true) used [OK]

Common Mistakes:

• Accessing array as object after json_decode(true)
• Assuming JSON string is invalid
• Forgetting to pass true for array output

5. You have a PHP array:

$data = ["id" => 101, "tags" => ["php", "json", "coding"]];

Which code correctly encodes this array to JSON and then decodes it back to a PHP object?

hard

A. $json = json_encode($data);
$obj = json_decode($json, true);

B. $json = json_decode($data);
$obj = json_encode($json);

C. $json = json_encode($data);
$obj = json_decode($json);

D. $json = json_encode($data, true);
$obj = json_decode($json, true);

Solution

1. Step 1: Encode PHP array to JSON string

   Use json_encode($data) to convert the PHP array into a JSON string.

2. Step 2: Decode JSON string back to PHP object

   Calling json_decode($json) without the second parameter returns a PHP object.

3. Final Answer:

   $json = json_encode($data); $obj = json_decode($json); -> Option C

4. Quick Check:

   Encode then decode without true returns object [OK]

Hint: json_decode() without true returns object [OK]

Common Mistakes:

• Using json_decode before json_encode
• Passing true to json_encode (invalid)
• Decoding with true returns array, not object