Working with JSON files in C Sharp (C#) - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is JSON and why is it commonly used in programming?

JSON (JavaScript Object Notation) is a simple text format to store and exchange data. It is easy for humans to read and write, and easy for machines to parse and generate. It is widely used for data exchange between web servers and clients.

beginner

Which C# class is commonly used to read and write JSON files?

The <code>System.Text.Json.JsonSerializer</code> class is commonly used in C# to convert objects to JSON strings and JSON strings back to objects. It helps read from and write to JSON files easily.

intermediate

How do you read a JSON file into a C# object?

You can read a JSON file by first reading the file content as a string using File.ReadAllText, then use JsonSerializer.Deserialize<T>(jsonString) to convert it into a C# object of type T.

intermediate

How do you write a C# object to a JSON file?

Use JsonSerializer.Serialize(object) to convert the object to a JSON string, then write it to a file using File.WriteAllText. This saves the object data in JSON format.

intermediate

What is the importance of matching C# class structure with JSON data?

The C# class properties must match the JSON keys to correctly convert JSON data to objects and vice versa. If they don't match, data may be lost or cause errors during serialization or deserialization.

Which method reads the entire content of a JSON file as a string in C#?

A File.ReadLine

B File.ReadAllText

C JsonSerializer.Deserialize

D JsonSerializer.Serialize

File.ReadAllText reads the whole file content as a string, which can then be deserialized.

Which class is used to convert JSON strings to C# objects?

A JsonReader

B File

C StreamReader

D JsonSerializer

JsonSerializer provides methods to serialize and deserialize JSON data.

What happens if the C# class property names do not match JSON keys during deserialization?

A Properties with no match are ignored

B Program crashes immediately

C JSON file is deleted

D All data is loaded correctly

Properties without matching JSON keys are ignored, so data may be missing.

How do you save a C# object as a JSON file?

A Serialize object to string, then write string to file

B Use File.ReadAllText

C Use JsonSerializer.Deserialize

D Use Console.WriteLine

You serialize the object to a JSON string and then write that string to a file.

Which namespace must you include to work with JSON serialization in C#?

A System.Net

B System.IO

C System.Text.Json

D System.Collections

System.Text.Json contains the JsonSerializer class for JSON operations.

Explain the steps to read data from a JSON file and convert it into a C# object.

Think about file reading and deserialization.

Describe how to save a C# object into a JSON file.

Focus on serialization and file writing.

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