public void TestResolvableEnumeration()
{
DictionaryResolvable resolvable = new DictionaryResolvable
{
{
"A", new List <IResolvable>
{
new DictionaryResolvable
{
{ "A", "A" },
{ "B", "B" },
{ "C", "C" },
},
new DictionaryResolvable
{
{ "A", "D" },
{ "B", "E" },
{ "C", "F" },
},
new DictionaryResolvable
{
{ "A", "G" },
{ "B", "H" },
{ "C", "J" },
{ "<INDEX>", "I" }
}
}
}
};
FormatBuilder builder =
new FormatBuilder().AppendFormat("{0:{A:[{<ITEMS>:{<INDEX>} {A} {B} {C}}{<JOIN>:, }]}}", resolvable);
Assert.AreEqual("[0 A B C, 1 D E F, I G H J]", builder.ToString());
}
public void TestNestedResolver()
{
DictionaryResolvable resolvable = new DictionaryResolvable
{
{
"A", new DictionaryResolvable
{
{"A", "aa"},
{"B", "ab"}
}
},
{
"B", new DictionaryResolvable
{
{"A", "ba"},
{"B", "bb"}
}
}
};
FormatBuilder builder = new FormatBuilder().AppendFormat("{0:{A:{B:{A}}} {B:{A:{B}}}}", resolvable);
Assert.AreEqual("aa bb", builder.ToString());
/* WHY????
* Step 1: {0:...} is resolved to 'resolvable'
* Step 2: There's a format, so the resolvable is called to resolve it.
* Step 3: {A:...} and {B:...} are resolved to two new Resolvables (the two inside the dictionary above).
* This bits, important, these are the last two resolvables on the stack...
* Step 4: {B:{A}} is resolved to "ab" (note not a resolvable), and {A:{B}} is resolved to "ba".
* Step 5: {A} is resolved by the resolvable on the stack at that point (the one belonging to 'A') as the B
* is not a resolvable itself. Therefore the A is "aa". Similarly for the {B}.
*
* {Context:\r\n{Key}\t: {Value}} =>
* {Context.Resource:\r\n{Key}\t: {Value}
*/
}
public void TestNestedResolver()
{
DictionaryResolvable resolvable = new DictionaryResolvable
{
{
"A", new DictionaryResolvable
{
{ "A", "aa" },
{ "B", "ab" }
}
},
{
"B", new DictionaryResolvable
{
{ "A", "ba" },
{ "B", "bb" }
}
}
};
FormatBuilder builder = new FormatBuilder().AppendFormat("{0:{A:{B:{A}}} {B:{A:{B}}}}", resolvable);
Assert.AreEqual("aa bb", builder.ToString());
/* WHY????
* Step 1: {0:...} is resolved to 'resolvable'
* Step 2: There's a format, so the resolvable is called to resolve it.
* Step 3: {A:...} and {B:...} are resolved to two new Resolvables (the two inside the dictionary above).
* This bits, important, these are the last two resolvables on the stack...
* Step 4: {B:{A}} is resolved to "ab" (note not a resolvable), and {A:{B}} is resolved to "ba".
* Step 5: {A} is resolved by the resolvable on the stack at that point (the one belonging to 'A') as the B
* is not a resolvable itself. Therefore the A is "aa". Similarly for the {B}.
*
* {Context:\r\n{Key}\t: {Value}} =>
* {Context.Resource:\r\n{Key}\t: {Value}
*/
}
public void TestResolvableEnumeration()
{
DictionaryResolvable resolvable = new DictionaryResolvable
{
{
"A", new List<IResolvable>
{
new DictionaryResolvable
{
{"A", "A"},
{"B", "B"},
{"C", "C"},
},
new DictionaryResolvable
{
{"A", "D"},
{"B", "E"},
{"C", "F"},
},
new DictionaryResolvable
{
{"A", "G"},
{"B", "H"},
{"C", "J"},
{"<INDEX>", "I"}
}
}
}
};
FormatBuilder builder =
new FormatBuilder().AppendFormat("{0:{A:[{<ITEMS>:{<INDEX>} {A} {B} {C}}{<JOIN>:, }]}}", resolvable);
Assert.AreEqual("[0 A B C, 1 D E F, I G H J]", builder.ToString());
}
