How to map multiple source members so that propertyMap.SourceMembers yields these multiple source members?









up vote
1
down vote

favorite












I want to reuse the intialized mappings of the c# Automapper (using version 6.2.2) by looping through all mapped properties.



Let's suppose I have the following:



public class Person

public int Id get; set;
public string FirstName get; set;
public string LastName get; set;


public class PersonDto

public int Id get; set;
public string FullName get; set;


AutoMapper.Mapper.Initialize(cfg =>

cfg.CreateMap<Person, PersonDto>()
.ForMember(dest => dest.FullName,
opt => opt.MapFrom(src => src.FirstName + " " + src.LastName));
;


The loop for Id behaves as expected:



var map = AutoMapper.Mapper.Configuration.FindTypeMapFor<Person, PersonDto>();
foreach (var propertyMap in map.GetPropertyMaps())

var destProp = propertyMap.DestinationProperty.Name; // = "Id"
var sourceMember = propertyMap.SourceMember.Name; // = "Id"
var sourceMembers = propertyMap.SourceMembers; // Count = 1



But when I loop through the FullName property mapping, I want to achieve that the propertyMap.SourceMembers results in the two SourceMembers FirstName and LastName:



var map = AutoMapper.Mapper.Configuration.FindTypeMapFor<Person, PersonDto>();
foreach (var propertyMap in map.GetPropertyMaps())

var destProp = propertyMap.DestinationProperty.Name; // = "FullName"
var sourceMember = propertyMap.SourceMember.Name; // = "LastName" (I don't care)
var sourceMembers = propertyMap.SourceMembers; // Count = 0 (want to achieve 2 for FirstName and LastName)



My goal is to create an automated similar mapping for an orderBy functionality based on the initialized automapper mappings. So I need to know (a) the order and (b) the sourceMembers. In the above case I want get the mapping for FullName from its source members FirstName and LastName (in this order).



Is it somehow possible to correctly register multiple source members so that propertyMap.SourceMembers yields all mapped source members? If yes, what should the map initialization look like?



PS: I don't want to write the orderBy mappings by hand, since I already have mappings thanks to automapper.










share|improve this question

















  • 1




    github.com/AutoMapper/AutoMapper/blob/…
    – Lucian Bargaoanu
    Nov 10 at 11:48










  • I didn't find smth that does exactly what you need, but that's the idea. Have a visitor that accumulates MemberExpression-s.
    – Lucian Bargaoanu
    Nov 10 at 16:18










  • Thank you for the link Lucian Bargaoanu, I think I know where to go now :-) I'll post my answer as soon as I have a robust solution
    – philipp-fx
    Nov 10 at 16:28














up vote
1
down vote

favorite












I want to reuse the intialized mappings of the c# Automapper (using version 6.2.2) by looping through all mapped properties.



Let's suppose I have the following:



public class Person

public int Id get; set;
public string FirstName get; set;
public string LastName get; set;


public class PersonDto

public int Id get; set;
public string FullName get; set;


AutoMapper.Mapper.Initialize(cfg =>

cfg.CreateMap<Person, PersonDto>()
.ForMember(dest => dest.FullName,
opt => opt.MapFrom(src => src.FirstName + " " + src.LastName));
;


The loop for Id behaves as expected:



var map = AutoMapper.Mapper.Configuration.FindTypeMapFor<Person, PersonDto>();
foreach (var propertyMap in map.GetPropertyMaps())

var destProp = propertyMap.DestinationProperty.Name; // = "Id"
var sourceMember = propertyMap.SourceMember.Name; // = "Id"
var sourceMembers = propertyMap.SourceMembers; // Count = 1



But when I loop through the FullName property mapping, I want to achieve that the propertyMap.SourceMembers results in the two SourceMembers FirstName and LastName:



var map = AutoMapper.Mapper.Configuration.FindTypeMapFor<Person, PersonDto>();
foreach (var propertyMap in map.GetPropertyMaps())

var destProp = propertyMap.DestinationProperty.Name; // = "FullName"
var sourceMember = propertyMap.SourceMember.Name; // = "LastName" (I don't care)
var sourceMembers = propertyMap.SourceMembers; // Count = 0 (want to achieve 2 for FirstName and LastName)



My goal is to create an automated similar mapping for an orderBy functionality based on the initialized automapper mappings. So I need to know (a) the order and (b) the sourceMembers. In the above case I want get the mapping for FullName from its source members FirstName and LastName (in this order).



Is it somehow possible to correctly register multiple source members so that propertyMap.SourceMembers yields all mapped source members? If yes, what should the map initialization look like?



PS: I don't want to write the orderBy mappings by hand, since I already have mappings thanks to automapper.










share|improve this question

















  • 1




    github.com/AutoMapper/AutoMapper/blob/…
    – Lucian Bargaoanu
    Nov 10 at 11:48










  • I didn't find smth that does exactly what you need, but that's the idea. Have a visitor that accumulates MemberExpression-s.
    – Lucian Bargaoanu
    Nov 10 at 16:18










  • Thank you for the link Lucian Bargaoanu, I think I know where to go now :-) I'll post my answer as soon as I have a robust solution
    – philipp-fx
    Nov 10 at 16:28












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I want to reuse the intialized mappings of the c# Automapper (using version 6.2.2) by looping through all mapped properties.



Let's suppose I have the following:



public class Person

public int Id get; set;
public string FirstName get; set;
public string LastName get; set;


public class PersonDto

public int Id get; set;
public string FullName get; set;


AutoMapper.Mapper.Initialize(cfg =>

cfg.CreateMap<Person, PersonDto>()
.ForMember(dest => dest.FullName,
opt => opt.MapFrom(src => src.FirstName + " " + src.LastName));
;


The loop for Id behaves as expected:



var map = AutoMapper.Mapper.Configuration.FindTypeMapFor<Person, PersonDto>();
foreach (var propertyMap in map.GetPropertyMaps())

var destProp = propertyMap.DestinationProperty.Name; // = "Id"
var sourceMember = propertyMap.SourceMember.Name; // = "Id"
var sourceMembers = propertyMap.SourceMembers; // Count = 1



But when I loop through the FullName property mapping, I want to achieve that the propertyMap.SourceMembers results in the two SourceMembers FirstName and LastName:



var map = AutoMapper.Mapper.Configuration.FindTypeMapFor<Person, PersonDto>();
foreach (var propertyMap in map.GetPropertyMaps())

var destProp = propertyMap.DestinationProperty.Name; // = "FullName"
var sourceMember = propertyMap.SourceMember.Name; // = "LastName" (I don't care)
var sourceMembers = propertyMap.SourceMembers; // Count = 0 (want to achieve 2 for FirstName and LastName)



My goal is to create an automated similar mapping for an orderBy functionality based on the initialized automapper mappings. So I need to know (a) the order and (b) the sourceMembers. In the above case I want get the mapping for FullName from its source members FirstName and LastName (in this order).



Is it somehow possible to correctly register multiple source members so that propertyMap.SourceMembers yields all mapped source members? If yes, what should the map initialization look like?



PS: I don't want to write the orderBy mappings by hand, since I already have mappings thanks to automapper.










share|improve this question













I want to reuse the intialized mappings of the c# Automapper (using version 6.2.2) by looping through all mapped properties.



Let's suppose I have the following:



public class Person

public int Id get; set;
public string FirstName get; set;
public string LastName get; set;


public class PersonDto

public int Id get; set;
public string FullName get; set;


AutoMapper.Mapper.Initialize(cfg =>

cfg.CreateMap<Person, PersonDto>()
.ForMember(dest => dest.FullName,
opt => opt.MapFrom(src => src.FirstName + " " + src.LastName));
;


The loop for Id behaves as expected:



var map = AutoMapper.Mapper.Configuration.FindTypeMapFor<Person, PersonDto>();
foreach (var propertyMap in map.GetPropertyMaps())

var destProp = propertyMap.DestinationProperty.Name; // = "Id"
var sourceMember = propertyMap.SourceMember.Name; // = "Id"
var sourceMembers = propertyMap.SourceMembers; // Count = 1



But when I loop through the FullName property mapping, I want to achieve that the propertyMap.SourceMembers results in the two SourceMembers FirstName and LastName:



var map = AutoMapper.Mapper.Configuration.FindTypeMapFor<Person, PersonDto>();
foreach (var propertyMap in map.GetPropertyMaps())

var destProp = propertyMap.DestinationProperty.Name; // = "FullName"
var sourceMember = propertyMap.SourceMember.Name; // = "LastName" (I don't care)
var sourceMembers = propertyMap.SourceMembers; // Count = 0 (want to achieve 2 for FirstName and LastName)



My goal is to create an automated similar mapping for an orderBy functionality based on the initialized automapper mappings. So I need to know (a) the order and (b) the sourceMembers. In the above case I want get the mapping for FullName from its source members FirstName and LastName (in this order).



Is it somehow possible to correctly register multiple source members so that propertyMap.SourceMembers yields all mapped source members? If yes, what should the map initialization look like?



PS: I don't want to write the orderBy mappings by hand, since I already have mappings thanks to automapper.







c# automapper






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 10 at 11:37









philipp-fx

71111




71111







  • 1




    github.com/AutoMapper/AutoMapper/blob/…
    – Lucian Bargaoanu
    Nov 10 at 11:48










  • I didn't find smth that does exactly what you need, but that's the idea. Have a visitor that accumulates MemberExpression-s.
    – Lucian Bargaoanu
    Nov 10 at 16:18










  • Thank you for the link Lucian Bargaoanu, I think I know where to go now :-) I'll post my answer as soon as I have a robust solution
    – philipp-fx
    Nov 10 at 16:28












  • 1




    github.com/AutoMapper/AutoMapper/blob/…
    – Lucian Bargaoanu
    Nov 10 at 11:48










  • I didn't find smth that does exactly what you need, but that's the idea. Have a visitor that accumulates MemberExpression-s.
    – Lucian Bargaoanu
    Nov 10 at 16:18










  • Thank you for the link Lucian Bargaoanu, I think I know where to go now :-) I'll post my answer as soon as I have a robust solution
    – philipp-fx
    Nov 10 at 16:28







1




1




github.com/AutoMapper/AutoMapper/blob/…
– Lucian Bargaoanu
Nov 10 at 11:48




github.com/AutoMapper/AutoMapper/blob/…
– Lucian Bargaoanu
Nov 10 at 11:48












I didn't find smth that does exactly what you need, but that's the idea. Have a visitor that accumulates MemberExpression-s.
– Lucian Bargaoanu
Nov 10 at 16:18




I didn't find smth that does exactly what you need, but that's the idea. Have a visitor that accumulates MemberExpression-s.
– Lucian Bargaoanu
Nov 10 at 16:18












Thank you for the link Lucian Bargaoanu, I think I know where to go now :-) I'll post my answer as soon as I have a robust solution
– philipp-fx
Nov 10 at 16:28




Thank you for the link Lucian Bargaoanu, I think I know where to go now :-) I'll post my answer as soon as I have a robust solution
– philipp-fx
Nov 10 at 16:28












1 Answer
1






active

oldest

votes

















up vote
0
down vote













Thanks to Lucian Bargaoanu, I developed the following dirty workaround.



Please note that this is the first time that I get in touch with interpreting expressions, so bear with me the quality of my solution:



var map = AutoMapper.Mapper.Configuration.FindTypeMapFor<Person, PersonDto>();
foreach (var propertyMap in map.PropertyMaps)

var destProp = propertyMap.DestinationMember.Name; // = "FullName"
var sourceMember = propertyMap.SourceMember?.Name; // = "LastName" (I don't care)
var sourceMembers = propertyMap.SourceMembers.ToList();
if (sourceMembers.Count() == 0)

ResolveSourceMembersInOrder(sourceMembers,
propertyMap.CustomMapExpression.Body as BinaryExpression);


// sourceMembers now yield the two desired sourceMembers "FirstName" and "LastName"


public static void ResolveSourceMembersInOrder(List<MemberInfo> memberInfos, BinaryExpression expression)

if (memberInfos == null)

memberInfos = new List<MemberInfo>();


if (expression == null)

return;


var left = expression.Left;
if (left.NodeType == ExpressionType.MemberAccess)

memberInfos.Add(MemberVisitor.GetMemberPath(left).FirstOrDefault());

else

ResolveSourceMembersInOrder(memberInfos, left as BinaryExpression);


var right = expression.Right;
if (right.NodeType == ExpressionType.MemberAccess)

memberInfos.Add(MemberVisitor.GetMemberPath(right).FirstOrDefault());

else

ResolveSourceMembersInOrder(memberInfos, right as BinaryExpression);







share|improve this answer
















  • 1




    See this. It returns things in the right order, but a few tests wouldn't hurt :)
    – Lucian Bargaoanu
    Nov 11 at 9:50










Your Answer






StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53238544%2fhow-to-map-multiple-source-members-so-that-propertymap-sourcemembers-yields-thes%23new-answer', 'question_page');

);

Post as a guest






























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













Thanks to Lucian Bargaoanu, I developed the following dirty workaround.



Please note that this is the first time that I get in touch with interpreting expressions, so bear with me the quality of my solution:



var map = AutoMapper.Mapper.Configuration.FindTypeMapFor<Person, PersonDto>();
foreach (var propertyMap in map.PropertyMaps)

var destProp = propertyMap.DestinationMember.Name; // = "FullName"
var sourceMember = propertyMap.SourceMember?.Name; // = "LastName" (I don't care)
var sourceMembers = propertyMap.SourceMembers.ToList();
if (sourceMembers.Count() == 0)

ResolveSourceMembersInOrder(sourceMembers,
propertyMap.CustomMapExpression.Body as BinaryExpression);


// sourceMembers now yield the two desired sourceMembers "FirstName" and "LastName"


public static void ResolveSourceMembersInOrder(List<MemberInfo> memberInfos, BinaryExpression expression)

if (memberInfos == null)

memberInfos = new List<MemberInfo>();


if (expression == null)

return;


var left = expression.Left;
if (left.NodeType == ExpressionType.MemberAccess)

memberInfos.Add(MemberVisitor.GetMemberPath(left).FirstOrDefault());

else

ResolveSourceMembersInOrder(memberInfos, left as BinaryExpression);


var right = expression.Right;
if (right.NodeType == ExpressionType.MemberAccess)

memberInfos.Add(MemberVisitor.GetMemberPath(right).FirstOrDefault());

else

ResolveSourceMembersInOrder(memberInfos, right as BinaryExpression);







share|improve this answer
















  • 1




    See this. It returns things in the right order, but a few tests wouldn't hurt :)
    – Lucian Bargaoanu
    Nov 11 at 9:50














up vote
0
down vote













Thanks to Lucian Bargaoanu, I developed the following dirty workaround.



Please note that this is the first time that I get in touch with interpreting expressions, so bear with me the quality of my solution:



var map = AutoMapper.Mapper.Configuration.FindTypeMapFor<Person, PersonDto>();
foreach (var propertyMap in map.PropertyMaps)

var destProp = propertyMap.DestinationMember.Name; // = "FullName"
var sourceMember = propertyMap.SourceMember?.Name; // = "LastName" (I don't care)
var sourceMembers = propertyMap.SourceMembers.ToList();
if (sourceMembers.Count() == 0)

ResolveSourceMembersInOrder(sourceMembers,
propertyMap.CustomMapExpression.Body as BinaryExpression);


// sourceMembers now yield the two desired sourceMembers "FirstName" and "LastName"


public static void ResolveSourceMembersInOrder(List<MemberInfo> memberInfos, BinaryExpression expression)

if (memberInfos == null)

memberInfos = new List<MemberInfo>();


if (expression == null)

return;


var left = expression.Left;
if (left.NodeType == ExpressionType.MemberAccess)

memberInfos.Add(MemberVisitor.GetMemberPath(left).FirstOrDefault());

else

ResolveSourceMembersInOrder(memberInfos, left as BinaryExpression);


var right = expression.Right;
if (right.NodeType == ExpressionType.MemberAccess)

memberInfos.Add(MemberVisitor.GetMemberPath(right).FirstOrDefault());

else

ResolveSourceMembersInOrder(memberInfos, right as BinaryExpression);







share|improve this answer
















  • 1




    See this. It returns things in the right order, but a few tests wouldn't hurt :)
    – Lucian Bargaoanu
    Nov 11 at 9:50












up vote
0
down vote










up vote
0
down vote









Thanks to Lucian Bargaoanu, I developed the following dirty workaround.



Please note that this is the first time that I get in touch with interpreting expressions, so bear with me the quality of my solution:



var map = AutoMapper.Mapper.Configuration.FindTypeMapFor<Person, PersonDto>();
foreach (var propertyMap in map.PropertyMaps)

var destProp = propertyMap.DestinationMember.Name; // = "FullName"
var sourceMember = propertyMap.SourceMember?.Name; // = "LastName" (I don't care)
var sourceMembers = propertyMap.SourceMembers.ToList();
if (sourceMembers.Count() == 0)

ResolveSourceMembersInOrder(sourceMembers,
propertyMap.CustomMapExpression.Body as BinaryExpression);


// sourceMembers now yield the two desired sourceMembers "FirstName" and "LastName"


public static void ResolveSourceMembersInOrder(List<MemberInfo> memberInfos, BinaryExpression expression)

if (memberInfos == null)

memberInfos = new List<MemberInfo>();


if (expression == null)

return;


var left = expression.Left;
if (left.NodeType == ExpressionType.MemberAccess)

memberInfos.Add(MemberVisitor.GetMemberPath(left).FirstOrDefault());

else

ResolveSourceMembersInOrder(memberInfos, left as BinaryExpression);


var right = expression.Right;
if (right.NodeType == ExpressionType.MemberAccess)

memberInfos.Add(MemberVisitor.GetMemberPath(right).FirstOrDefault());

else

ResolveSourceMembersInOrder(memberInfos, right as BinaryExpression);







share|improve this answer












Thanks to Lucian Bargaoanu, I developed the following dirty workaround.



Please note that this is the first time that I get in touch with interpreting expressions, so bear with me the quality of my solution:



var map = AutoMapper.Mapper.Configuration.FindTypeMapFor<Person, PersonDto>();
foreach (var propertyMap in map.PropertyMaps)

var destProp = propertyMap.DestinationMember.Name; // = "FullName"
var sourceMember = propertyMap.SourceMember?.Name; // = "LastName" (I don't care)
var sourceMembers = propertyMap.SourceMembers.ToList();
if (sourceMembers.Count() == 0)

ResolveSourceMembersInOrder(sourceMembers,
propertyMap.CustomMapExpression.Body as BinaryExpression);


// sourceMembers now yield the two desired sourceMembers "FirstName" and "LastName"


public static void ResolveSourceMembersInOrder(List<MemberInfo> memberInfos, BinaryExpression expression)

if (memberInfos == null)

memberInfos = new List<MemberInfo>();


if (expression == null)

return;


var left = expression.Left;
if (left.NodeType == ExpressionType.MemberAccess)

memberInfos.Add(MemberVisitor.GetMemberPath(left).FirstOrDefault());

else

ResolveSourceMembersInOrder(memberInfos, left as BinaryExpression);


var right = expression.Right;
if (right.NodeType == ExpressionType.MemberAccess)

memberInfos.Add(MemberVisitor.GetMemberPath(right).FirstOrDefault());

else

ResolveSourceMembersInOrder(memberInfos, right as BinaryExpression);








share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 10 at 17:26









philipp-fx

71111




71111







  • 1




    See this. It returns things in the right order, but a few tests wouldn't hurt :)
    – Lucian Bargaoanu
    Nov 11 at 9:50












  • 1




    See this. It returns things in the right order, but a few tests wouldn't hurt :)
    – Lucian Bargaoanu
    Nov 11 at 9:50







1




1




See this. It returns things in the right order, but a few tests wouldn't hurt :)
– Lucian Bargaoanu
Nov 11 at 9:50




See this. It returns things in the right order, but a few tests wouldn't hurt :)
– Lucian Bargaoanu
Nov 11 at 9:50

















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53238544%2fhow-to-map-multiple-source-members-so-that-propertymap-sourcemembers-yields-thes%23new-answer', 'question_page');

);

Post as a guest














































































這個網誌中的熱門文章

Barbados

How to read a connectionString WITH PROVIDER in .NET Core?

Node.js Script on GitHub Pages or Amazon S3