Working with JSON files in C Sharp (C#) - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to read a JSON file into a string.

C Sharp (C#)

string jsonString = File.[1]("data.json");

Drag options to blanks, or click blank then click option'

AWriteAllLines

BWriteAllText

CReadAllLines

DReadAllText

Attempts:

3 left

💡 Hint

Common Mistakes

Using WriteAllText instead of ReadAllText will overwrite the file.

ReadAllLines returns a string array, not a single string.

The File.ReadAllText method reads the entire content of a file as a string, which is perfect for reading JSON files.

2fill in blank

medium

Complete the code to deserialize JSON string into an object of type Person.

C Sharp (C#)

Person person = JsonSerializer.[1]<Person>(jsonString);

Drag options to blanks, or click blank then click option'

AParse

BDeserialize

CSerialize

DConvert

Attempts:

3 left

💡 Hint

Common Mistakes

Using Serialize instead of Deserialize will convert an object to JSON, not the other way.

Parse is not a method in JsonSerializer.

The JsonSerializer.Deserialize<T> method converts a JSON string into an object of type T.

3fill in blank

hard

Fix the error in the code to serialize an object to JSON string.

C Sharp (C#)

string jsonString = JsonSerializer.[1](person);

Drag options to blanks, or click blank then click option'

ADeserialize

BParse

CSerialize

DConvert

Attempts:

3 left

💡 Hint

Common Mistakes

Using Deserialize will cause a compile-time error because it expects a JSON string input.

Parse and Convert are not methods of JsonSerializer.

The Serialize method converts an object into a JSON string.

4fill in blank

hard

Fill both blanks to write a JSON string to a file asynchronously.

C Sharp (C#)

await File.[1]Async("output.json", jsonString[2]);

Drag options to blanks, or click blank then click option'

AWriteAllText

BReadAllText

CWriteAllLines

D, CancellationToken.None

Attempts:

3 left

💡 Hint

Common Mistakes

Using ReadAllTextAsync instead of WriteAllTextAsync will read instead of write.

Omitting the cancellation token parameter may cause errors in some contexts.

The WriteAllTextAsync method writes text to a file asynchronously. The second blank is a cancellation token parameter, which can be CancellationToken.None if no cancellation is needed.

5fill in blank

hard

Fill all three blanks to create a dictionary from JSON and filter keys with length > 3.

C Sharp (C#)

var filtered = jsonDict.Where(kv => kv.Key.[1] > 3).ToDictionary(kv => kv.[2], kv => kv.[3]);

Drag options to blanks, or click blank then click option'

ALength

BKey

CValue

DCount

Attempts:

3 left

💡 Hint

Common Mistakes

Using Count instead of Length for string length.

Swapping Key and Value in ToDictionary parameters.

We use Length to get the length of the key string, then select Key and Value to build the filtered dictionary.

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