本文整理汇总了C#中SourceCode.GetFileName方法的典型用法代码示例。如果您正苦于以下问题：C# SourceCode.GetFileName方法的具体用法？C# SourceCode.GetFileName怎么用？C# SourceCode.GetFileName使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类SourceCode的用法示例。
在下文中一共展示了SourceCode.GetFileName方法的4个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C#代码示例。
示例1: Compile
public TargetCodeResult Compile(SourceCode aSourceCode, string aCurrentCodeLine, int aLinePosition)
{
TargetCodeResult result = new TargetCodeResult(aCurrentCodeLine);
String usingNamespace = GetUsingNamespace(aSourceCode, aCurrentCodeLine);
bool correctLine = IsUsingCorrect(usingNamespace);
if (!correctLine && (IsSystemNamespace(usingNamespace) || IsProgramNamespace(aSourceCode, usingNamespace)))
return new TargetCodeResult("");
if (correctLine)
{
// TODO: For now we import all the Java classes in that package, we don't worry much about that since the Java compiler will take care of this for us
// Suggest that we do change this in the future if required
StringBuilder newLine = new StringBuilder();
newLine.Append("import ").Append(usingNamespace).Append(".*;");
result = new TargetCodeResult(newLine.ToString());
}
else
{
StringBuilder newLine = new StringBuilder();
newLine.Append("//");
newLine.Append(aCurrentCodeLine);
newLine.Append(" // Not supported yet");
result = new TargetCodeResult(newLine.ToString());
result.LogError(aSourceCode.GetFileName() + ": Using directive not supported yet on line: " + aLinePosition);
}
return result;
}示例2: DoValidation
public String DoValidation(SourceCode aSourceCode)
{
String result = null;
int totalClasses = CountClassName(aSourceCode);
int totalInterfaces = CountInterfaces(aSourceCode);
if (totalClasses == 0 && totalInterfaces == 0)
result = aSourceCode.GetFileName() + ": A C# file must at least contain a class or an interface. ";
if (totalClasses == 1 && totalInterfaces != 0)
result = aSourceCode.GetFileName() + ": A C# file can only have one class or interface just like Java. ";
else if (totalInterfaces == 1 && totalClasses != 0)
result = aSourceCode.GetFileName() + ": A C# file can only have one class or interface just like Java. ";
return result;
}示例3: Identify
public bool Identify(SourceCode aSourceCode, string aCurrentCodeLine, int aLinePosition)
{
// TODO: We need to also cater for C# properties later on
bool result = false;
if (aSourceCode.CountTokens(aCurrentCodeLine, '(') == 1 &&
aSourceCode.CountTokens(aCurrentCodeLine, ')') == 1 &&
aCurrentCodeLine.IndexOf(aSourceCode.GetFileName()) == -1 &&
!aSourceCode.ContainKeyword(aCurrentCodeLine, "class") &&
aCurrentCodeLine.Split(' ').Length > 2 &&
!MainMethodComp.IdentifyMainMethod(aSourceCode, aCurrentCodeLine, aLinePosition) &&
(aCurrentCodeLine.EndsWith("{") || aSourceCode.GetNextLine(aLinePosition).StartsWith("{")))
result = true;
return result;
}示例4: DoValidation
public string DoValidation(SourceCode aSourceCode)
{
String result = null;
bool foundIndexers = false;
int pos = 1;
foreach (String line in aSourceCode.GetLines())
{
if (line.IndexOf("this[") > -1 && (line.EndsWith("{") || aSourceCode.GetNextLine(pos).StartsWith("{")))
foundIndexers = true;
pos++;
}
if (foundIndexers)
result = aSourceCode.GetFileName() + ": Java unfortunatly doesn't support C# indexers. ";
return result;
}