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How to use char.TryParse with a string


Since a string may be an arbitrary amount of characters, you cannot implicitly cast it to a char. You would need to either change this part input.Key.ToString() to input.KeyChar or do something like this input.Key.ToString().ToCharArray()[0]


Categories : C#

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 TryParse. 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.

Categories : C#
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(",","."); Now // 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 invariant cul

Categories : C#
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.

Categories : Java
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

Categories : Java
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"); should be 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

Categories : C#
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