Re: Why does this not change my string? [modified] - C# Discussion Boards

Re: Why does this not change my string?

PIEBALDconsult15-Aug-11 16:01

PIEBALDconsult

15-Aug-11 16:01

You're passing Mel Brooks by reference? OMG | :OMG:

Re: Why does this not change my string?

Luc Pattyn15-Aug-11 16:30

Luc Pattyn

15-Aug-11 16:30

of course, there's too much value to pass on.

Smile | :)

Luc Pattyn [My Articles] Nil Volentibus Arduum

Re: Why does this not change my string?

fcronin15-Aug-11 12:08

fcronin

15-Aug-11 12:08

private string encodeString(string input)
{
   string output = string.Empty;

   for (int i = 0; i < input.Length-1; i++)
   {
      if ((i + 1) % 6 == 0)
      {
         output = input.Insert(i, "-");
      }
   }

   return output;
}

Re: Why does this not change my string?

stephen.darling15-Aug-11 12:14

stephen.darling

15-Aug-11 12:14

Hello.

I tried your code, however, it produces the following result...

"38459171915879268-601"

So, it appears to be adding the whole string and then a '-', instead of a '-' every 5 charictures?

Thank you,
Stephen

Re: Why does this not change my string?

fcronin16-Aug-11 2:39

fcronin

16-Aug-11 2:39

I apologize... I threw that in haste without really thinking it through... this is probably the easiest way to handle any length string...

static string encodeString(string input)
{
    string output = string.Empty;
    for (int i = 0; i < input.Length; i++)
    {
        output += input.Substring(i, 1);
        if ((i+1) % 6 == 0 && i != (input.Length-1))
            output += "-";
    }
    return output;
}

Re: Why does this not change my string?

stephen.darling16-Aug-11 7:27

stephen.darling

16-Aug-11 7:27

fcronin wrote:
I apologize... I threw that in haste without really thinking it through... this is probably the easiest way to handle any length string...

No problem!
Thank you again,

Steve

Re: Why does this not change my string?

PIEBALDconsult15-Aug-11 15:30

PIEBALDconsult

15-Aug-11 15:30

And you can increment i by six if you like; you're not limited to incrementing by one.

Re: Why does this not change my string? [modified]

Shameel15-Aug-11 21:33

Shameel

15-Aug-11 21:33

stephen.darling wrote:
It simply does not alter the string.

And it will not, no matter what you do. As others have already pointed out, strings are immutable in .NET which means they cannot be altered once they're created.

Take a look at this example:

string a = "abc";
a = a + "xyz";

You might be thinking that in line 2 the characters "xyz" is appended to the string a, but that is not the case. When line 2 is executed, the Runtime creates a new string that is the result of the concatenation i.e. "abcxyz" and makes the variable a point to the new string. The old "abc" string becomes garbage (since it has no valid reference) and is collected during the next GC cycle.

To make large changes to to strings, it is recommended to use the System.Text.StringBuilder class.

private string encodeString(string input)
{
    StringBuilder builder = new StringBuilder();
    for (int i = 0; i < input.Length; i++) {
        builder.Append(input[i]);
        if ((i + 1) % 6 == 0) {
            builder.Append("-");
        }
    }
    return builder.ToString();
}

And use your method like this:

string input = "00000000000000000000";
string output = encodeString(input); //output contains "00000-00000-00000-00000"

modified on Tuesday, August 16, 2011 2:45 PM

Re: Why does this not change my string?

Matt Meyer16-Aug-11 5:50

Matt Meyer

16-Aug-11 5:50

i < input.Length-1

I think you mean

i < input.Length

Otherwise you're dropping the last character.

Re: Why does this not change my string?

Shameel16-Aug-11 8:45

Shameel

16-Aug-11 8:45

Correct. Thanks for pointing out. Answer updated Smile | :)

Find asp.net listview control in Item Template

akosidandan15-Aug-11 6:05

akosidandan

15-Aug-11 6:05

Hello Experts,

I would like to ask how will I find the listview control or use the correct event. My goal is that I want to check the listview control text value inside of listview item template after it load the data, then I have a condition to trigger some code depends on what my control text result.

Here is my Code I am not sure why it does not work and it always give me

Object reference not set to an instance of an object.

The aspx page Code

XML

<ItemTemplate>
               <tr style="">
                   <td>
                       <asp:Label ID="AnnouncementMessageLabel" runat="server"
                           Text='<%# Eval("AnnouncementMessage") %>' />
                   </td>
                   <td>
                       <asp:Literal ID="projectType" runat="server"
                       Text='<%# Eval("AnnouncementType") %>'>
                       </asp:Literal>
                       <asp:Button ID="Button1" runat="server" Text="Button" />
                   </td>
               </tr>
</ItemTemplate>

My behind file code.

protected void ListView1_DataBound(object sender, EventArgs e)
        {
            foreach (ListViewDataItem item in ListView1.Items)
            {
                Literal project = new Literal();
                Button b = new Button();
                b = (Button)item.FindControl("Button1");
                project =(Literal)item.FindControl("projectType");
                if (project.Text == "Project")
                {
                    b.Text = "This is a project POST";
                }

            }

        }

Any comments or suggestions are kindly appreciated.

Re: Find asp.net listview control in Item Template

akosidandan15-Aug-11 7:29

akosidandan

15-Aug-11 7:29

Fix it when I almost lost hope trying Big Grin | :-D

when I suddenly copy paste my design of item template to alternating item template of my listview. It seems that the listview databound event is affected by both alternating item template and item template.

Deserialise complicated Jason data

AndieDu14-Aug-11 15:55

AndieDu

14-Aug-11 15:55

Dear all,

I have this jason data that ready to be serialised: (Indention added for readability)

JavaScript

{"Error":"", 
"UserMeasurements": 
        [{        "MeasurementName":"BloodPressure", 
                  "ClinicalDatas": 
                        [{"Name":"Systolic", 
                          "Unit":"mmHg      ", 
                          "ClinicalDataPoints":[[1304197968110,149],[1304290213920,145],[1304369712717,159],[1304459900650,160], 
                                                                        [1304542964530,176],[1304564292340,147],[1304649916767,141],[1304726125653,139],[1304826583873,156], 
                                                                        [1304880946500,167],[1305069333890,152]]}, 
                         {"Name":"Diastolic", 
                          "Unit":"mmHg      ", 
                          "ClinicalDataPoints":[[1304197968220,129],[1304290215047,92],[1304369712763,91],[1304459900900,105], 
                                                                        [1304542965530,83],[1304564292403,103],[1304649916797,81],[1304726125687,133],[1304826584000,85], 
                                                                        [1304880946547,108],[1305069334890,89]]}, 
                         {"Name":"Mean Blood Pressure", 
                          "Unit":"mmHg      ", 
                          "ClinicalDataPoints":[]} 
                        ] 
                } 
        ] 
}";

and I have these c# code to deserialised it:

namespace Json 
{ 
    public partial class _Default : System.Web.UI.Page 
    { 
        protected void Page_Load(object sender, EventArgs e) 
        { 
            JavaScriptSerializer js = new JavaScriptSerializer(); 
            string test = "{\"Error\":\"\",\"UserMeasurements\":[{\"MeasurementName\":\"BloodPressure\",\"ClinicalDatas\":[{\"Name\":\"Systolic\",\"Unit\":\"mmHg      \",\"ClinicalDataPoints\":[[1304197968110,149],[1304290213920,145],[1304369712717,159],[1304459900650,160],[1304542964530,176],[1304564292340,147],[1304649916767,141],[1304726125653,139],[1304826583873,156],[1304880946500,167],[1305069333890,152]]},{\"Name\":\"Diastolic\",\"Unit\":\"mmHg      \",\"ClinicalDataPoints\":[[1304197968220,129],[1304290215047,92],[1304369712763,91],[1304459900900,105],[1304542965530,83],[1304564292403,103],[1304649916797,81],[1304726125687,133],[1304826584000,85],[1304880946547,108],[1305069334890,89]]},{\"Name\":\"Mean Blood Pressure\",\"Unit\":\"mmHg      \",\"ClinicalDataPoints\":[]}]}]}"; 
 
            Measurement result = js.Deserialize<Measurement>(test); 
        } 
    } 
 
 
    public class Measurement 
    { 
        public string Error { get; set; } 
        public List<UserMeasurement> UserMeasurements { get; set; } 
    } 
 
    public class UserMeasurement 
    { 
        public string MeasurementName { get; set; } 
        public List<ClinicalDatas> UserClinicalDatas { get; set; } 
    } 
 
    public class ClinicalDatas : List<Measurement> 
    { 
        public string Name { get; set; } 
        public string Unit { get; set; } 
        public string[][] ClinicalDataPoints { get; set; } 
    } 
}

I am able to get the value for MeasurementName property, but the UserClinicalDatas always returns null, can someone in here point me out why this is the case? Many thanks.

Re: Deserialise complicated Jason data

BobJanova14-Aug-11 23:18

BobJanova

14-Aug-11 23:18

The property appears to be called 'ClinicalDatas' (not UserClinicalDatas) in the JSON.

Regarding the names, 'data' is already plural, so using 'datas' just looks wrong. If you are in charge of both ends of this serialisation you should change that.

Re: Deserialise complicated Jason data

AndieDu15-Aug-11 14:42

AndieDu

15-Aug-11 14:42

right on the spot, Bob, and amazingly, i didnt spot out this subtle thing, thanks heaps anyway.

Re: Deserialise complicated Jason data

Bernhard Hiller17-Aug-11 1:00

Bernhard Hiller

17-Aug-11 1:00

Try a tool for generating your classes from the Json data, e.g. http://json2csharp.com/[^]. That will prevent typos.

C# MySQL Question

d87c14-Aug-11 13:18

d87c

14-Aug-11 13:18

Hi All, I have a problem

I have a MySQL Table, one of the columns' Data Type is TinyInt(1)

When I use this line of code it gives an error

int newNumber = int.Parse(string.Format("{0}", reader["someField"]));

I think the reader is reading a bool value

however MySQL Table has row values of 0 & 1 & 6 & 7

so it is not bool true/false type

How can I actually grab that value to store into "int newNumber"

Thanks

Re: C# MySQL Question

Richard Andrew x6414-Aug-11 13:44

Richard Andrew x64

14-Aug-11 13:44

The reader returns type object, and it must be cast to the necessary type for use in function calls.

For the TinyInt type, I forget exactly what you need to cast it to but you can try short.

Try this:

int newNumber = (int)(short)reader["someField"];

The difficult we do right away...
...the impossible takes slightly longer.

Re: C# MySQL Question

d87c14-Aug-11 15:33

d87c

14-Aug-11 15:33

thank for your reply, however, I got an error and it says Specific Cast is not valid

Re: C# MySQL Question

Richard Andrew x6414-Aug-11 15:39

Richard Andrew x64

14-Aug-11 15:39

What about:

int newNumber = (int)(byte)reader["fieldName"];

The difficult we do right away...
...the impossible takes slightly longer.

Re: C# MySQL Question

d87c14-Aug-11 16:57

d87c

14-Aug-11 16:57

I tried byte as well, still same error, specific cast .... =(

Re: C# MySQL Question

PIEBALDconsult14-Aug-11 17:59

PIEBALDconsult

14-Aug-11 17:59

How about an sbyte then?

Message Removed

15-Aug-11 2:44

marssilen

15-Aug-11 2:44

Message Removed

Re: lame-ass hack attempt removed

#realJSOP15-Aug-11 2:50

#realJSOP

15-Aug-11 2:50

Nice try, retard.

What makes you think a web site by and for programmers is not going to have that base covered?

Oh, and by the way, your account will be deleted in the next 30 minutes or so.

".45 ACP - because shooting twice is just silly" - JSOP, 2010
-----
You can never have too much ammo - unless you're swimming, or on fire. - JSOP, 2010
-----
"Why don't you tie a kerosene-soaked rag around your ankles so the ants won't climb up and eat your candy ass." - Dale Earnhardt, 1997

Re: lame-ass hack attempt removed

OriginalGriff15-Aug-11 3:05

OriginalGriff

15-Aug-11 3:05

Normally, I do not agree with your occasional use of "retard". In this case however, it is accurate and justified. 5! Laugh | :laugh:

Real men don't use instructions. They are only the manufacturers opinion on how to put the thing together.

Manfred R. Bihy: "Looks as if OP is learning resistant."

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