how to get 3rd row or 2nd row values from ienumerable object.

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hi all,
i do have a requirement that to retrive data which is available in the form of ienumerable<someclassname>, in that i am getting 5 rows of data, from them i need to bind 1st row values to some asp.net textboxes, and 2nd row values to some other textboxes, and 3rd row values to some other.
for that i have written the code as follows.

i am getting the required data from businessclass library in the form of "ienumerable<benefeciarydetails>" type.

Ienumerable<beneficiarydetails>[] requireddata=new Ienumerable<beneficiarydetails>[] { getBeneficiarydetails()};

so, here to treive and assigning data to required textboxes i have written as follows.

for(var i=0;i<requireddata.length;i++){>
if(i==0){
txtlname.text=requireddata[i].lastname;
txtmiddlename.text=requiredata[i].middlename;
}
if(i==1){
txtlnamesecond.text=requireddata[i].lastname;
txtmiddlenamesecond.text=requireddata[i].middlename;
}
if(i==2){
txtlnamethird.text=requrieddata[i].lastname;
txtmiddlenamethired.text=requireddata[i].middlename;
}
.
.
.
.
}

here if the requireddata contains only one row then i need to assign those data to 1st required textboxes. and for all empty will be placed. i have tried above code, but not getting proper result, plz help me to solve this, or should i change my code.

Posted 9-Feb-15 6:26am

pavan_ kumar

Updated 9-Feb-15 6:27am

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Richard Deeming · Answer 1 · 2015-02-09T06:36:00

The problem is that you've declared an array of IEnumerable instances, which contains a single element - the results of your getBeneficiarydetails method.

Assuming the getBeneficiarydetails method actually returns an IEnumerable<SomeType>, where SomeType is a class which contains properties called lastname and middlename, then something like this should work:

C#

IEnumerable<SomeType> requiredData = getBeneficiarydetails();
int index = 0;

foreach (SomeType item in requiredData)
{
    switch (index)
    {
        case 0:
        {
            txtlname.Text = item.lastname;
            txtmiddlename.Text = item.middlename;
            break;
        }
        case 1:
        {
            txtlnamesecond.Text = item.lastname;
            txtmiddlenamesecond.Text = item.middlename;
            break;
        }
        ...
    }
    
    index++;
}