Copy Item's Metadata for batching - msbuild

Question

This is an extension to this question.
Suppose I'm creating items from other item's metadata:
<ItemGroup>
<A>
<files>f1;f2;f3;...</files>
<x>...</x>
<y>...</y>
<z>...</z>
...
</A>
<B Include="%(A.files)">
<x>%(A.x)</x>
<y>%(A.y)</y>
<z>%(A.z)</z>
...
</B>
</ItemGroup>
%(A.files) is a list of files separated by ;, such that for each A item I'm creating many B items (one for each file).
But frequently when I process B item I need the original's A item metadata. In this example I copied each metadata manually from A to B.
Is there a way to copy all of A's metadata to B without explicitly specifying each one of them?

Answer 1

Little late, but I like this solution better:
<B Include="#(A->Metadata('files'))" />
Full Example:
<?xml version="1.0" encoding="utf-8"?>
<Project DefaultTargets="Build" ToolsVersion="4.0" xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
<ItemGroup>
<A Include="A1">
<files>a1_file1.htm;a1_file2.htm</files>
<x>a1</x>
<y>b1</y>
<z>c1</z>
</A>
<A Include="A2">
<files>a2_file.proj</files>
<x>a2</x>
<y>b2</y>
<z>c2</z>
</A>
<B Include="#(A->Metadata('files'))" />
</ItemGroup>
<Target Name="Build">
<Message Text="A: #(A->'%(Identity) x:%(x) y:%(y) z:%(z) files:%(files)', '
')" />
<Message Text="B: #(B->'%(Identity) x:%(x) y:%(y) z:%(z) files:%(files)', '
')" />
</Target>
</Project>
Output:
Build:
A: A1 x:a1 y:b1 z:c1 files:a1_file1.htm;a1_file2.htm
A2 x:a2 y:b2 z:c2 files:a2_file.proj
B: a1_file1.htm x:a1 y:b1 z:c1 files:a1_file1.htm;a1_file2.htm
a1_file2.htm x:a1 y:b1 z:c1 files:a1_file1.htm;a1_file2.htm
a2_file.proj x:a2 y:b2 z:c2 files:a2_file.proj

Answer 2

As far as I can tell, MSBuild only copies metadata when there is a one-to-one mapping between the input and output item lists. In your case, you're starting with one item and expanding to many items. To get around this, I suggest using item batching:
<?xml version="1.0" encoding="iso-8859-1"?>
<Project
xmlns="http://schemas.microsoft.com/developer/msbuild/2003"
ToolsVersion="4.0"
DefaultTargets="Print">
<ItemGroup>
<A Include="A1">
<files>test.htm;test_sol1.htm</files>
<x>a1</x>
<y>b1</y>
<z>c1</z>
</A>
<A Include="A2">
<files>test.proj</files>
<x>a2</x>
<y>b2</y>
<z>c2</z>
</A>
</ItemGroup>
<Target Name="ExpandA">
<ItemGroup>
<ExpandedA Include="%(A.files)">
<Original>%(Identity)</Original>
</ExpandedA>
</ItemGroup>
</Target>
<Target
Name="CopyMetadata"
Outputs="%(ExpandedA.Identity)"
DependsOnTargets="ExpandA">
<PropertyGroup>
<ExpandedAIdentity>%(ExpandedA.Identity)</ExpandedAIdentity>
<ExpandedAOriginal>%(ExpandedA.Original)</ExpandedAOriginal>
</PropertyGroup>
<ItemGroup>
<ExpandedAMetadata Include="#(A)" Condition=" '%(Identity)' == '$(ExpandedAOriginal)' ">
<Expanded>$(ExpandedAIdentity)</Expanded>
</ExpandedAMetadata>
</ItemGroup>
</Target>
<Target Name="Print" DependsOnTargets="CopyMetadata">
<ItemGroup>
<B Include="#(ExpandedAMetadata->'%(Expanded)')" />
</ItemGroup>
<!--Use commas to illustrate that "files" has been expanded-->
<Message Text="A: %(A.files)" />
<Message Text="ExpandedA: #(ExpandedA, ',')" />
<Message Text="ExpandedAMetadata: #(ExpandedAMetadata, ',')" />
<Message Text="B: #(B->'%(Identity) x:%(x) y:%(y) z:%(z)', ',')" />
</Target>
</Project>
and the output of the "Print" target:
Print:
A: test.htm;test_sol1.htm
A: test.proj
ExpandedA: test.htm,test_sol1.htm,test.proj
ExpandedAMetadata: A1,A1,A2
B: test.htm x:a1 y:b1 z:c1,test_sol1.htm x:a1 y:b1 z:c1,test.proj x:a2 y:b2 z:c2
ExpandedA is similar to B in your original question; it is the expanded version of A but without any metadata. Then I run the CopyMetadata target once for each item in ExpandedA (thanks to item batching). Each run, the original A item is copied to the ExpandedAMetadata item group along with all of its metadata. The Original metadata is used to ensure that the correct A item is associated with each file. Finally, in the Print target, B is constructed using an item transformation, so that all the metadata from ExpandedAMetadata is copied over as well.

Answer 3

I'm not sure if I get what you need, but you can copy all A metedata to B as in the answer for related question:
<B Include="#(A)">
It should copy all metedata from A to B.

Answer 4

Here is a direct answer to the original question:
<!-- Copy all metadata from an input item to an output item -->
<UsingTask
TaskName="CopyMetadataTask"
TaskFactory="CodeTaskFactory"
AssemblyFile="$(MSBuildToolsPath)\Microsoft.Build.Tasks.v4.0.dll" >
<ParameterGroup>
<InputItem ParameterType="Microsoft.Build.Framework.ITaskItem" Required="true" />
<OutputItem ParameterType="Microsoft.Build.Framework.ITaskItem" Required="true" />
</ParameterGroup>
<Task>
<Using Namespace="System"/>
<Code Type="Fragment" Language="cs">
<![CDATA[
InputItem.CopyMetadataTo(OutputItem);
]]>
</Code>
</Task>
</UsingTask>
You can use this task to copy metadata between items the way it was originally intended, and it can easily be adapted to multiple items on input/output, though doing so does induce issues whereby it is necessary to match the inputs and outputs 1/1, or make other compromises, though it can easily be adapted to the exact situation since you can modify the C# code however is necessary. E.g.:
<UsingTask
TaskName="CopyMetadataTask"
TaskFactory="CodeTaskFactory"
AssemblyFile="$(MSBuildToolsPath)\Microsoft.Build.Tasks.v4.0.dll" >
<ParameterGroup>
<InputItem ParameterType="Microsoft.Build.Framework.ITaskItem[]" Required="true" />
<OutputItem ParameterType="Microsoft.Build.Framework.ITaskItem[]" Required="true" />
</ParameterGroup>
<Task>
<Using Namespace="System"/>
<Code Type="Fragment" Language="cs">
<![CDATA[
for (int i = 0; i < InputItem.Length; ++i)
{
InputItem[i].CopyMetadataTo(OutputItem[i]);
}
]]>
</Code>
</Task>
</UsingTask>
The task can then be called to copy metadata directly, for example (using the second version):
<ItemGroup>
<ItemListA Include="a;b;c" Metadata="MetadataValue" />
<ItemListB Include="d;e;f" OtherMetadata="OtherValue" />
</ItemGroup>
<CopyMetadataTask InputItem="#(ItemListA)" OutputItem="#(ItemListB)" />