We know that in an appropriate setting, with some exact sequence $0 \to A \to B \to C \to 0$ of $R$-modules with suitable $R$ (commutative ring with unity), we have injective resolution of $A,C$. Then, the Horseshoe lemma says that we have injective resolution of $B$ which makes every diagram commutes. Then, we can take $H^{n}$ functor to get a long exact sequence.
My question is, if we know injective resolution of $B$ and $C$, which already commutes, is it possible to get an injective resolution of $A$ which commutes with injective resolution of $B$? Furthermore, from the long exact sequence of cohomology, can we calculate $H^{n}(A)$ from the information of $H^{n-1}(C)$ and $H^{n}(B)$?