Enable specifying output assembly name in compiler options
Addresses #1797
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8 changed files with 70 additions and 10 deletions
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@ -74,20 +74,20 @@ namespace Microsoft.DotNet.Tools.Compiler.Tests
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[Fact]
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public void LibraryWithAnalyzer()
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{
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{
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var root = Temp.CreateDirectory();
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var testLibDir = root.CreateDirectory("TestLibraryWithAnalyzer");
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var sourceTestLibDir = Path.Combine(_testProjectsRoot, "TestLibraryWithAnalyzer");
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CopyProjectToTempDir(sourceTestLibDir, testLibDir);
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// run compile
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var outputDir = Path.Combine(testLibDir.Path, "bin");
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var testProject = GetProjectPath(testLibDir);
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var buildCmd = new BuildCommand(testProject, output: outputDir, framework: DefaultFramework);
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var result = buildCmd.ExecuteWithCapturedOutput();
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result.Should().Pass();
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Assert.Contains("CA1018", result.StdErr);
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}
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@ -103,7 +103,7 @@ namespace Microsoft.DotNet.Tools.Compiler.Tests
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var testProjectDir = Path.Combine(_testProjectsRoot, "TestAppCompilationContext", "TestApp");
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var testProject = Path.Combine(testProjectDir, "project.json");
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var buildCommand = new BuildCommand(testProject);
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buildCommand.Execute().Should().Pass();
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@ -167,6 +167,24 @@ namespace Microsoft.DotNet.Tools.Compiler.Tests
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runCommand.Execute().Should().Pass();
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}
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[Fact]
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public void CanSetOutputAssemblyName()
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{
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var testInstance =
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TestAssetsManager
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.CreateTestInstance("LibraryWithOutputAssemblyName")
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.WithLockFiles();
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var root = testInstance.TestRoot;
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var outputDir = Path.Combine(root, "bin");
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var testProject = ProjectUtils.GetProjectJson(root, "LibraryWithOutputAssemblyName");
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var buildCommand = new BuildCommand(testProject, output: outputDir, framework: DefaultFramework);
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var result = buildCommand.ExecuteWithCapturedOutput();
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result.Should().Pass();
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new DirectoryInfo(outputDir).Should().HaveFiles(new [] { "MyLibrary.dll" });
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}
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private void CopyProjectToTempDir(string projectDir, TempDirectory tempDir)
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{
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// copy all the files to temp dir
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