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Autofixture - base property in new list on property from a previously constructed one

How to create a new object instance from a TypeHow to get the list of properties of a class?Using LINQ to remove elements from a List<T>Get property value from string using reflection in C#How to Sort a List<T> by a property in the objectHow do I create a list of classes which always have a predefined value set in AutoFixture?Preventing Autofixture from filling child collectionsWhy not inherit from List<T>?Autofixture, create a list of dictionary<string, object> with alternating keyAutoFixture sometimes failing to generate a value for a property marked with RangeAttribute

1

I think I'm missing something, but what I'm trying to do is this:

I have two database entities represented in my C# code. One is a child of the other, therefore the child contains a field which should reference the parent's ID.

The parent class is the following

public class Product

 public Guid Id get; set; 

 public string Name get; set; 

 public string Description get; set; 

 public decimal Price get; set; 

 public decimal DeliveryPrice get; set; 

The child class is the following:

public class ProductOption

 public Guid Id get; set; 

 public Guid ProductId get; set; 

 public string Name get; set; 

 public string Description get; set; 

I have created a list of random "parents" as such:

var products = fixture.CreateMany<Product>(5).ToList();

What I wanted to do was then create 10 child objects, and randomly give them a ProductId from the list of products created by AutoFixture. So I tried this:

var rand = new Random();
var options = fixture.Build<ProductOption>()
 .With(option => option.ProductId, products[rand.Next(0, 5)].Id)
 .CreateMany(10)
 .ToList();

It almost worked, but what I found was that all the ProductIds were the same one, so it obviously only ever hit rand.Next once.

Is what I'm doing even possible/advisable?

c# .net unit-testing testing autofixture

share|improve this question

asked Mar 7 at 23:47

Thanasi PoulosThanasi Poulos

148114

add a comment | 

1

I think I'm missing something, but what I'm trying to do is this:

I have two database entities represented in my C# code. One is a child of the other, therefore the child contains a field which should reference the parent's ID.

The parent class is the following

public class Product

 public Guid Id get; set; 

 public string Name get; set; 

 public string Description get; set; 

 public decimal Price get; set; 

 public decimal DeliveryPrice get; set; 

The child class is the following:

public class ProductOption

 public Guid Id get; set; 

 public Guid ProductId get; set; 

 public string Name get; set; 

 public string Description get; set; 

I have created a list of random "parents" as such:

var products = fixture.CreateMany<Product>(5).ToList();

What I wanted to do was then create 10 child objects, and randomly give them a ProductId from the list of products created by AutoFixture. So I tried this:

var rand = new Random();
var options = fixture.Build<ProductOption>()
 .With(option => option.ProductId, products[rand.Next(0, 5)].Id)
 .CreateMany(10)
 .ToList();

It almost worked, but what I found was that all the ProductIds were the same one, so it obviously only ever hit rand.Next once.

Is what I'm doing even possible/advisable?

c# .net unit-testing testing autofixture

share|improve this question

asked Mar 7 at 23:47

Thanasi PoulosThanasi Poulos

148114

add a comment | 

I think I'm missing something, but what I'm trying to do is this:

I have two database entities represented in my C# code. One is a child of the other, therefore the child contains a field which should reference the parent's ID.

The parent class is the following

public class Product

 public Guid Id get; set; 

 public string Name get; set; 

 public string Description get; set; 

 public decimal Price get; set; 

 public decimal DeliveryPrice get; set; 

The child class is the following:

public class ProductOption

 public Guid Id get; set; 

 public Guid ProductId get; set; 

 public string Name get; set; 

 public string Description get; set; 

I have created a list of random "parents" as such:

var products = fixture.CreateMany<Product>(5).ToList();

What I wanted to do was then create 10 child objects, and randomly give them a ProductId from the list of products created by AutoFixture. So I tried this:

var rand = new Random();
var options = fixture.Build<ProductOption>()
 .With(option => option.ProductId, products[rand.Next(0, 5)].Id)
 .CreateMany(10)
 .ToList();

It almost worked, but what I found was that all the ProductIds were the same one, so it obviously only ever hit rand.Next once.

Is what I'm doing even possible/advisable?

c# .net unit-testing testing autofixture

share|improve this question

asked Mar 7 at 23:47

Thanasi PoulosThanasi Poulos

148114

public class Product

 public Guid Id get; set; 

 public string Name get; set; 

 public string Description get; set; 

 public decimal Price get; set; 

 public decimal DeliveryPrice get; set; 

public class ProductOption

 public Guid Id get; set; 

 public Guid ProductId get; set; 

 public string Name get; set; 

 public string Description get; set; 

var products = fixture.CreateMany<Product>(5).ToList();

var rand = new Random();
var options = fixture.Build<ProductOption>()
 .With(option => option.ProductId, products[rand.Next(0, 5)].Id)
 .CreateMany(10)
 .ToList();

share|improve this question

asked Mar 7 at 23:47

Thanasi PoulosThanasi Poulos

148114

share|improve this question

asked Mar 7 at 23:47

Thanasi PoulosThanasi Poulos

148114

asked Mar 7 at 23:47

Thanasi PoulosThanasi Poulos

148114

asked Mar 7 at 23:47

Thanasi PoulosThanasi Poulos

148114

1 Answer
1

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oldest

votes

3

When I provide value for the property, I would expect all instances build with same builder/fixture will have provided value.

So what you noticed is desired behaviour.

Instead of already generated value you can provide a "factory" which will generate value for the property during instance creation.

Latest Autofixture version introduced overload for .With method which accepts a function as parameter.

var rand = new Random();
Func<Guid> pickProductId = () => products[rand.Next(0, 5)].Id;

var options = 
 fixture.Build<ProductOption>()
 .With(option => option.ProductId, pickProductId)
 .CreateMany(10)
 .ToList();

// Prove
options.Select(o => o.ProductId).ToHashSet().Should().HaveCountGreaterThan(1); // Pass Ok

share|improve this answer

answered Mar 8 at 0:06

FabioFabio

20.1k22048

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ah yes, I was looking through the intellisense overloads but I got a little confused. This actually makes perfect sense, a function that serves as a "generator" Thank you!
  
  – Thanasi Poulos
  Mar 8 at 0:08
  
  

  
  
  
  

add a comment | 

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3

When I provide value for the property, I would expect all instances build with same builder/fixture will have provided value.

So what you noticed is desired behaviour.

Instead of already generated value you can provide a "factory" which will generate value for the property during instance creation.

Latest Autofixture version introduced overload for .With method which accepts a function as parameter.

var rand = new Random();
Func<Guid> pickProductId = () => products[rand.Next(0, 5)].Id;

var options = 
 fixture.Build<ProductOption>()
 .With(option => option.ProductId, pickProductId)
 .CreateMany(10)
 .ToList();

// Prove
options.Select(o => o.ProductId).ToHashSet().Should().HaveCountGreaterThan(1); // Pass Ok

share|improve this answer

answered Mar 8 at 0:06

FabioFabio

20.1k22048

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ah yes, I was looking through the intellisense overloads but I got a little confused. This actually makes perfect sense, a function that serves as a "generator" Thank you!
  
  – Thanasi Poulos
  Mar 8 at 0:08
  
  

  
  
  
  

add a comment | 

3

When I provide value for the property, I would expect all instances build with same builder/fixture will have provided value.

So what you noticed is desired behaviour.

Instead of already generated value you can provide a "factory" which will generate value for the property during instance creation.

Latest Autofixture version introduced overload for .With method which accepts a function as parameter.

var rand = new Random();
Func<Guid> pickProductId = () => products[rand.Next(0, 5)].Id;

var options = 
 fixture.Build<ProductOption>()
 .With(option => option.ProductId, pickProductId)
 .CreateMany(10)
 .ToList();

// Prove
options.Select(o => o.ProductId).ToHashSet().Should().HaveCountGreaterThan(1); // Pass Ok

share|improve this answer

answered Mar 8 at 0:06

FabioFabio

20.1k22048

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ah yes, I was looking through the intellisense overloads but I got a little confused. This actually makes perfect sense, a function that serves as a "generator" Thank you!
  
  – Thanasi Poulos
  Mar 8 at 0:08
  
  

  
  
  
  

add a comment | 

When I provide value for the property, I would expect all instances build with same builder/fixture will have provided value.

So what you noticed is desired behaviour.

Instead of already generated value you can provide a "factory" which will generate value for the property during instance creation.

Latest Autofixture version introduced overload for .With method which accepts a function as parameter.

var rand = new Random();
Func<Guid> pickProductId = () => products[rand.Next(0, 5)].Id;

var options = 
 fixture.Build<ProductOption>()
 .With(option => option.ProductId, pickProductId)
 .CreateMany(10)
 .ToList();

// Prove
options.Select(o => o.ProductId).ToHashSet().Should().HaveCountGreaterThan(1); // Pass Ok

share|improve this answer

answered Mar 8 at 0:06

FabioFabio

20.1k22048

var rand = new Random();
Func<Guid> pickProductId = () => products[rand.Next(0, 5)].Id;

var options = 
 fixture.Build<ProductOption>()
 .With(option => option.ProductId, pickProductId)
 .CreateMany(10)
 .ToList();

// Prove
options.Select(o => o.ProductId).ToHashSet().Should().HaveCountGreaterThan(1); // Pass Ok

share|improve this answer

answered Mar 8 at 0:06

FabioFabio

20.1k22048

answered Mar 8 at 0:06

FabioFabio

20.1k22048

answered Mar 8 at 0:06

FabioFabio

20.1k22048