Your group is such that $|G|=p^n$ and $|\Phi(G)|=p$. Since $(C_p)^{n-1}$ is completely reducible, there is a subgroup $H$ of $G$ such that $G=HZ(G)$ and $H\cap Z(G)=\Phi(G)$. Thus $H$ is an extra-special group (possibly trivial), and we are taking the central product with the abelian group $Z(G)$, which is either of the form $(C_p)^m$ or $(C_{p^2})\times(C_p)^m$. The first case is easy, since again, it is completely reducible, so we get a group of the form (extra-special) times (some copies of $C_p$). The second case also gives (some group) times (some copies of $C_p$). I believe the (some group) is uniquely determined by its order (that is the central product of either of the two non-abelian groups of order $p^3$ and $C_{p^2}$ are isomorphic), but I haven't checked any cases but $p=2$.

Steve