Let $F$ be a number field and $S$ be a finite set of places of $F$ including archimedian places. Let $\zeta^S(s)$ be the partial L-function, that is the meromorphic continuation of the product of local zeta functions $\zeta_v(s)=\frac{1}{1-q^{-s}}$ outside the places in $S$.
Then we know that the complete zeta function $\zeta(s)$ has a simple pole at $s=0$.
Since $\zeta(s)=\zeta^S(s)\cdot \prod_{v \in S}\zeta_v(s)$ for $\Re(s) \gg0$ and each $\zeta_(s),\zeta^S(s),\zeta_v(s)$ has meromorphic continuation to $\mathbb{C}$, I suppose the order of $\zeta^S(s)$ at $s=0$ is (the order of $\zeta(s)$ at $s=0$) - $\sum_{v \in S}ord_{s=0}(\zeta_v(s))$.
So I think $\zeta^S(s)$ would have a pole or have zero at $s=0$ depending the set of $S$.
My reasoning is right?