interpret string as variable in C#

Question

I have some public const strings in c# console application as shown below:

//Account one
    public const string POP_USER1 = "[email protected]";
    public const string POP_PWD1 = "abc";

    //Account two
    public const string POP_USER2 = "[email protected]";
    public const string POP_PWD2 = "xyz";

    //Account three
    public const string POP_USER3 = "[email protected]";
    public const string POP_PWD3 = "pqr;

We are using c# MailMan to retrieve emails present in those accounts. I simply wrote a for loop 3 times:

for (int i = 1; i <= 3; i++)
 {
    eEmails obj = new eEmails (i);
 }

In the constructor of eEmails, I am writing the below logic:

public eEmails (int counter)
{
   MailMan obj = new MailMan()
   obj.PopUsername =  "POP_USER" + counter;
   obj.PopPassword = "POP_PWD" + counter;
}

The lines where I am assigning user name and passwords, I need to fetch the exact const variable (i.e., POP_USER1, POP_USER2, POP_USER3 etc.,)

However I am not able to get the variable dynamically. I can simply write 3 if blocks in eEmails (int counter), but I didnt like that way. can somebody advise a better way of handling this situation without using separate if blocks for each user??

Define the data as Array or List? List<POPUserData> for example? — TGlatzer
– TGlatzer, Commented Jan 27, 2014 at 10:28
Defining your constants as Arrays of String(String[] POPUSER = new String[] {"User1","User2","User3"}; and the same for passwords) would do the trick. I think it is the simplest solution. Note, however, that defining the passwords unencrypted in the source code is not safe. — ThunderGr
– ThunderGr, Commented Jan 27, 2014 at 10:31
Your current eEmails implementation has a bug. You are assigning the password to the username property. — Tim Schmelter
– Tim Schmelter, Commented Jan 27, 2014 at 10:32
@ThunderGr: don't do like that. Your data structure implies that all usernames belong together and all password belong together. That's not a good idea. Instead a username and a password belongs together and you want a list of that. See Tim's answer. — Thomas Weller
– Thomas Weller, Commented Jan 27, 2014 at 10:58

Tim Schmelter · Accepted Answer · 2014-01-27 10:49:01Z

2

Use a class instead of strings, then your code becomes more redable and maintainable and it'll also be less error-prone. Here is an example using a List<PopServerAccount> as container:

public class PopServerAccount
{ 
    public string Username {get;set;}
    public string Password {get;set;}

    public override bool  Equals(object obj)
    {
        PopServerAccount p2 = obj as PopServerAccount;
        if (p2 == null) return false;
        return Username == p2.Username;
    }

    public override int GetHashCode()
    {
        return Username.GetHashCode();
    }

    public override string ToString()
    {
        return Username;
    }
}

now change the signature of your method:

public eEmails (PopServerAccount pop)
{
    MailMan obj = new MailMan()
    obj.PopUsername =  pop.Username;
    obj.PopPassword = pop.Password;
}

Sample data:

var myPopServers = new List<PopServerAccount> 
{
    new PopServerAccount{ Username = "[email protected]", Password = "abc"},new PopServerAccount{ Username = "[email protected]", Password = "xyz"}
};

Use a loop and call your method:

foreach (PopServerAccount pop in myPopServers)
{ 
    eEmails(pop);
}

edited Jan 27, 2014 at 10:49

answered Jan 27, 2014 at 10:30

Tim Schmelter

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5 Comments

mmssaann Over a year ago

not sure whats wrong, in eEmails(pop){} , I am not getting the properties of pop. when I say pop and put dot (.) intelligence not showing anything?

Tim Schmelter Over a year ago

@mmssaann: i cannot reproduce this issue. Maybe you need to show your code in your question.

mmssaann Over a year ago

Intelligence is not showing up only in that particular file, however the properties are accessible if I type them explicitly I am not getting any compile error. I think this is more of intelligence issue.

Tim Schmelter Over a year ago

@mmssaann: are these properties the only which you are missing or are no properties visible? Is this a separate class library, have you recompiled all (also in debug mode)?

mmssaann Over a year ago

none of the properties are visible I am afraid. All files are in the same application under same name space everything compiles successfully.

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