Displaying Image from File Attachment control in InfoPath form inside SSRS report.

I was assigned the task of displaying images attached to InfoPath form using file attachment control in one of the SQL Server Reporting Services Report.

For this we need to add an image control and set it’s property in the following manner

Image Source :- Database

Then select appropriate DataSet, Image Field and MimeType

Here to display the Image properly we need to write a Report Assembly with the following code.

public static class FileDecoder
    {
        /// <summary>
        /// Takes string value for image from the infopath form
        /// and converts it to proper bytes for displaying it as an image
        /// </summary>
        /// <param name="imageData">string representation of the image</param>
        /// <returns>corrected byte array</returns>
        public static byte[] GetCorrectedByte(string imageData)
        {
            byte[] attachmentNodeBytes = Convert.FromBase64String(imageData);
            // Position 20 contains a DWORD indicating the length of the
            // filename buffer. The filename is stored as Unicode so the
            // length is multiplied by 2.
            int fnLength = attachmentNodeBytes[20] * 2;
            // The file is located after the header, which is 24 bytes long
            // plus the length of the filename. 
            byte[] fileContents = new byte[attachmentNodeBytes.Length - (24 + fnLength)];

            for (int i = 0; i < fileContents.Length; ++i)
            {
                fileContents[i] = attachmentNodeBytes[24 + fnLength + i];
            }

            return fileContents;
        }

    }

 

And set the value for image control as

=(Namespace).FileDecoder.GetCorrectedByte(Fields!fieldName.Value)

 

Bye..

Using Left and CharIndex in Oracle

Well i was given the task to get the email id’s of the user which was stored in one of our oracle db table.

But the problem was that we wanted that part of emailid which  appears before ‘@’ .

Well coming from SQL Server background i thought it could be acheived using CharIndex and Left Function.

Let’s see what they do

Select Left(‘abcdef’,3)

-> abc

and

select charindex(‘c’,'abcde’)

-> 3

But than as expected there were no functions like charindex and left in Oracle.

After searching i finally managed to found the solution

Inplace of CharIndex we have instr function

select instr(‘ab’,'b’) from dual;

->2

and for left and also right we have

SUBSTR (`ABCDEF’,-5); //Right(..)

SUBSTR (`ABCDEF’,1,5); // Left(…

So finally the query was

substr(emailid,1,Instr(Emailid,’@')-1)

-1 is used otherwise @ will also come along

And one more thing, to extract username portion from login name i.e  nishantr1 from

abccompany\nishantr1 we could write something as following

SUBSTRING(loginname ,charindex(‘\’,loginname)+1, len(loginname)) for sql server.

Bye