|Related to : How to use char.TryParse with a string|
|How to use char.TryParse with my menu|
Console.ReadKey() returns a ConsoleKeyInfo instance. You can access
the char from that using the KeyChar property. No need to call
Your code doesn't work because ConsoleKeyInfo.ToString doesn't return
what you expect. According to the docs it:
Returns the fully qualified type name of this instance.
|How do I programm a language independent double.tryparse?|
10 years ago I would solve it like this (do not do that):
sDemo = sDemo.Replace(",",".");
// to string
var text = someDouble.ToString(NumberFormatInfo.InvariantInfo);
// and back
var number = double.Parse(text, NumberFormatInfo.InvariantInfo);
Key point is to use same culture. If you convert values for internal
use (to example, passing double value as a part of text) - use
|Why String created using new operator creates string literal in string pool|
First, I recommend that you not use new String("abc") because it
behaves as you described. Second, when you use new you should expect a
new Object instance will be created and it is.
|generic signature to accept Map or Map|
If you were only reading the values out of the Map you could have done
something like this:
public static Report createReport(Map<String,?> parameters)
However, since you also want to add enteries, you need to know the
type, or at least be able to know that the type in the Map is the same
as the type you are adding.
If it is variable, then I would suggest having another parameter t
|OracleParameterCollection.Add(string, type, string, size, string)|
You need to get rid of the colons in the Parameters.Add:
da.InsertCommand.Parameters.Add(":ID", OracleType.Number, 22, "ID");
da.InsertCommand.Parameters.Add("ID", OracleType.Number, 22, "ID");
The colon is used in the actual SQL string to denote a bind parameter.
When you create the parameter, just name without the colon.
See example here