MIPS Assembly, Merging Registers

I have a fairly quick question about MIPS programming. Assume there are 4 integers stored in 4 seperate registers (for arguments sake 1, 2, 3 and 4). How would one go about getting, in a single register, those merged together so inside the register was a single integer 1234. I heard mention of shifting bits to accomplish it but I can't fathom how shifting could help (since shifting is basically just multipling whatever integer is in the register by 2^k where k is the number of bytes shifted)

Any help would be amazing,
Tony V.

If your number is represented in Octal or Hexadecimal the shift and add can do the job otherwise, you are right, in a ten base you will have to
1234 = 1*10^3+2*10^2+3*10+4

all the best