Padding oracle attack, encrypting your own message My understanding of the encryption attack on a padding oracle is that the attacker has a set of $P_i$'s and $\operatorname{Dec}(C_i)$'s

I understand that by back calculating the attacker can find out $C_{i-1}$'s because we have

$$P_i = \operatorname{Dec}(C_i) \oplus C_{i-1}$$

This way the attacker can find all the $C_{i-1}$'s, but what happens with the very last $C_i$? Since you're starting from the end and going backwards, it looks like the very last $C_i$ wont get calculated, in which case the attacker isn't able to encrypt all of the message...?

EDIT: 

So i've run the decryption attack and obtained all the $Dec(C_i)$' for an intercepted ciphertext (so a bunch of blocks of $C_i$'s). and then using that i have calculated a sample $P_i$. I want to generate my own $P_i$ now and encrypt it. So what i've done is i take each block of my sample $P_i$ and XOR that with $Dec(C_{i-1})$ to get a C_i, however, the issue is that the response doesn't seem to be valid. So doesn't $Dec(C_i)$ directly depend on what $C_i$ is being passed into it? The conclusion i've made is that the $Dec(C_i)$ Value changes based on each block of ciphertext that's being fed into it, so the $C_i$'s that i generated using the previous $Dec(C_i)$'s won't be decrypted correctly, because the actual program computes a new set of $Dec(C_i)$'s and based on those my ciphertext isn't correct ... does that make sense? or am i misunderstanding the whole thing.
I'm assuming you mean CBC mode, and let's say PKCS7 padding. I haven't heard of this attack before, and can't really find any sources (please link if you have one! How would an attacker end up with the $\operatorname{Dec}(C_i)$'s?), but here's one way to generate a valid ciphertext.

If the attacker is okay with 15 garbage bytes at the end of his message, he could do the following:

Send $IV = C_{i-1}$ and a random ciphertext block $C_r$ to the oracle a bunch of times. For a little more than 1 in 256 times, $C_r$ will end up having valid padding (e.g. 0x01 for PKCS7). The attacker can simply add such a block $C_r$ to the end of his ciphertext ($C_1, \dotsc, C_{i-1}$) to get a valid encrypted message, albeit with (usually) 15 garbage bytes at the end of the decrypted, de-padded message.