Working with JSON files in C Sharp (C#) - Mini Project: Build & Apply

Working with JSON files

📖 Scenario: You are building a simple program to manage a list of books in a library. The data will be saved and loaded using JSON files.

🎯 Goal: Create a C# program that stores book information in a list, converts it to JSON format, and prints the JSON string.

📋 What You'll Learn

Create a list of books with specific properties

Add a configuration variable for JSON formatting

Serialize the list of books to a JSON string

Print the JSON string to the console

💡 Why This Matters

🌍 Real World

Many applications save and load data in JSON format because it is easy to read and write for both humans and computers.

💼 Career

Knowing how to work with JSON files is essential for software developers, especially when building web services, APIs, or data-driven applications.

1

Create the list of books

Create a list called books of type List<Book> and add these exact books: new Book { Title = "The Hobbit", Author = "J.R.R. Tolkien", Year = 1937 } and new Book { Title = "1984", Author = "George Orwell", Year = 1949 }. Also, define the Book class with public properties Title, Author, and Year.

C Sharp (C#)

// Define the Book class and create the books list
// Your code here

Hint

Define a class Book with three public properties. Then create a List<Book> and add two Book objects with the exact details.

2

Add JSON serialization options

Add a variable called options of type System.Text.Json.JsonSerializerOptions and set its WriteIndented property to true.

C Sharp (C#)

using System.Text.Json;

// Add JSON serialization options here
// Your code here

Hint

Use new JsonSerializerOptions { WriteIndented = true } to create the options variable.

3

Serialize the books list to JSON

Create a string variable called jsonString and assign it the JSON serialization of books using JsonSerializer.Serialize with options.

C Sharp (C#)

// Serialize the books list to JSON string
// Your code here

Hint

Use JsonSerializer.Serialize(books, options) to convert the list to a JSON string.

4

Print the JSON string

Write a line to print the jsonString variable to the console using Console.WriteLine.

C Sharp (C#)

// Print the JSON string to the console
// Your code here

Hint

Use Console.WriteLine(jsonString); to display the JSON output.

Practice

1. What is the main purpose of using JSON files in C# programming?

easy

A. To store and exchange data in a simple text format

B. To compile C# code faster

C. To create graphical user interfaces

D. To manage database connections

Solution

1. Step 1: Understand JSON file usage

   JSON files store data in a readable text format that can be shared or saved easily.

2. Step 2: Identify the correct purpose

   Among the options, only storing and exchanging data matches JSON's role.

3. Final Answer:

   To store and exchange data in a simple text format -> Option A

4. Quick Check:

   JSON = data storage and exchange [OK]

Hint: JSON files hold data as text for easy sharing [OK]

Common Mistakes:

• Thinking JSON compiles code
• Confusing JSON with UI design
• Assuming JSON manages databases

2. Which of the following is the correct way to read a JSON file content into a string in C#?

easy

A. string json = JsonSerializer.ReadFile("data.json");

B. string json = File.ReadAllText("data.json");

C. string json = File.ReadJson("data.json");

D. string json = JsonConvert.Read("data.json");

Solution

1. Step 1: Recall file reading method in C#

   The method File.ReadAllText reads all text from a file into a string.

2. Step 2: Check method correctness

   Only File.ReadAllText("data.json") is valid syntax to read JSON as string.

3. Final Answer:

   string json = File.ReadAllText("data.json"); -> Option B

4. Quick Check:

   File.ReadAllText reads file content [OK]

Hint: Use File.ReadAllText to read JSON file content [OK]

Common Mistakes:

• Using non-existent methods like ReadJson
• Confusing JsonSerializer with file reading
• Using JsonConvert without importing Newtonsoft

3. Given the following code, what will be the output?

using System.Text.Json;

var json = "{\"Name\":\"Alice\", \"Age\":30}";
var person = JsonSerializer.Deserialize<Person>(json);
Console.WriteLine(person.Name + " is " + person.Age + " years old.");

public class Person {
    public string Name { get; set; }
    public int Age { get; set; }
}

medium

A. Alice is 30 years old.

B. Name is Alice, Age is 30

C. System.Text.Json.JsonException

D. null is 0 years old.

Solution

1. Step 1: Deserialize JSON string to Person object

   The JsonSerializer converts the JSON string into a Person object with Name = "Alice" and Age = 30.

2. Step 2: Print the formatted string

   The Console.WriteLine outputs "Alice is 30 years old." by accessing person.Name and person.Age.

3. Final Answer:

   Alice is 30 years old. -> Option A

4. Quick Check:

   Deserialization + property access = output string [OK]

Hint: Deserialize JSON then access properties for output [OK]

Common Mistakes:

• Expecting JSON string printed directly
• Confusing property names or types
• Ignoring deserialization step

4. What is wrong with this code snippet that tries to write an object to a JSON file?

using System.Text.Json;
using System.IO;

var person = new Person { Name = "Bob", Age = 25 };
File.WriteAllText("person.json", person.ToString());

public class Person {
    public string Name { get; set; }
    public int Age { get; set; }
}

medium

A. File.WriteAllText cannot write to JSON files

B. Person class must be static to serialize

C. person.ToString() does not convert the object to JSON format

D. Missing using directive for System.IO

Solution

1. Step 1: Analyze how object is converted to JSON

   Calling person.ToString() returns the class name, not JSON text.

2. Step 2: Identify correct serialization method

   Use JsonSerializer.Serialize(person) to convert the object to JSON string before writing.

3. Final Answer:

   person.ToString() does not convert the object to JSON format -> Option C

4. Quick Check:

   To write JSON, serialize object first [OK]

Hint: Use JsonSerializer.Serialize, not ToString(), to get JSON [OK]

Common Mistakes:

• Using ToString() instead of serialization
• Assuming WriteAllText can't write JSON
• Thinking class must be static to serialize

5. You want to read a JSON file containing a list of products and convert it into a List<Product> in C#. Which code snippet correctly accomplishes this?

public class Product {
    public string Name { get; set; }
    public double Price { get; set; }
}

// JSON file content example: [{"Name":"Pen","Price":1.5},{"Name":"Book","Price":12.99}]

// Which code correctly reads and deserializes the JSON file?

hard

A. var products = JsonSerializer.Deserialize(File.ReadAllText("products.json"));
products = null;

B. var products = JsonSerializer.Deserialize("products.json");

C. var products = JsonSerializer.Deserialize(File.ReadAllLines("products.json"));

D. var json = File.ReadAllText("products.json");
var products = JsonSerializer.Deserialize(json);

Solution

1. Step 1: Read entire JSON file as string

   Use File.ReadAllText to get the JSON content from the file.

2. Step 2: Deserialize JSON string to List<Product>

   Use JsonSerializer.Deserialize(json) to convert JSON text into a list of Product objects.

3. Final Answer:

   var json = File.ReadAllText("products.json"); var products = JsonSerializer.Deserialize(json); -> Option D

4. Quick Check:

   ReadAllText + Deserialize = correct [OK]

Hint: Read file text first, then deserialize to list [OK]

Common Mistakes:

• Passing filename directly to Deserialize
• Using ReadAllLines instead of ReadAllText
• Setting deserialized list to null immediately