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Make LanguagePrimitive.GetEnumerable treat 'DataTable' as Enumerable (#6511)

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* Make LanguagePrimitive.GetEnumerable treat 'DataTable' as Enumerable
This commit is contained in:
Dongbo Wang 2018-03-29 16:22:44 -07:00 committed by Aditya Patwardhan
parent ec0f1116db
commit cd075ea4b4
2 changed files with 14 additions and 4 deletions
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4
src/System.Management.Automation/engine/LanguagePrimitives.cs
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@ -476,12 +476,10 @@ namespace System.Management.Automation
return null;
}
#if !CORECLR
private static IEnumerable DataTableEnumerable(object obj)
{
return (((DataTable)obj).Rows);
}
#endif
private static IEnumerable TypicalEnumerable(object obj)
{
@ -517,12 +515,10 @@ namespace System.Management.Automation
private static GetEnumerableDelegate CalculateGetEnumerable(Type objectType)
{
#if !CORECLR
if (typeof(DataTable).IsAssignableFrom(objectType))
{
return LanguagePrimitives.DataTableEnumerable;
}
#endif
// Don't treat IDictionary or XmlNode as enumerable...
if (typeof(IEnumerable).IsAssignableFrom(objectType)

14
test/powershell/engine/Api/LanguagePrimitive.Tests.ps1
View file

@ -32,4 +32,18 @@ Describe "Language Primitive Tests" -Tags "CI" {
$a[0] = $a = [PSObject](,1)
[System.Management.Automation.LanguagePrimitives]::IsTrue($a) | Should -BeTrue
}
It "LanguagePrimitives.GetEnumerable should treat 'DataTable' as Enumerable" {
$dt = [System.Data.DataTable]::new("test")
$dt.Columns.Add("Name", [string]) > $null
$dt.Columns.Add("Age", [string]) > $null
$dr = $dt.NewRow(); $dr["Name"] = "John"; $dr["Age"] = "20"
$dr2 = $dt.NewRow(); $dr["Name"] = "Susan"; $dr["Age"] = "25"
$dt.Rows.Add($dr); $dt.Rows.Add($dr2)
[System.Management.Automation.LanguagePrimitives]::IsObjectEnumerable($dt) | Should -BeTrue
$count = 0
[System.Management.Automation.LanguagePrimitives]::GetEnumerable($dt) | ForEach-Object { $count++ }
$count | Should -Be 2
}
}