Suppose $R$ is a Krull domain with the field of fraction $K$. To every prime ideal $P$ of $R$ of height $1$, one can associate a $ \mathbb{Z}$-valued discrete valuation which we denote by $v_P$.

Suppose $P_1, \dots, P_k$ are pairwise distinct height one prime ideas of $R$. It is well-known that the following approximation theorem holds: for any $(n_1, \dots, n_k) \in \mathbb{Z}^k$, there exists $x \in K$ such that

1. $v_{P_i} (x)= n_i$ for all $1 \le i \le k$
2. $v_{P}(x) \ge 0$ for all prime ideals $P$ of $R$ of height one other than $P_1, \dots, P_k$.

I would like to know if one can replace (2) with $v_{P}(x) = 0$ for all all prime ideals $P$ of height one other than $P_1, \dots, P_k$, perhaps at the cost of extending the initial set to a larger set of prime ideals.

In case it helps, the case I am really interested in is when $R$ is an integral extension of $ \mathbb{Q}[X_1, \dots, X_n]$. Any reference would be highly appreciated.

Edit: To clarify the question, what I would like to know if the following refinement of the approximation lemma holds:

Given $P_1, \dots, P_k$ are above, there exists a finite set $\mathcal{P}$ of prime ideals of height $1$ only depending on $P_1, \dots, P_k$ so that for every vector $(n_i)_{1 \le i \le k} \in \mathbb{Z}^k$, there exists $x \in K$ such that in addition to satisfying (1) and (2) above, it also satisfies $v_{P}(x)=0$ for all height one prime ideals $P$ outside $\mathcal{P}$.

I interpret the comment below by @Laurent Moret-Bailly as a negative answer to this. If this is indeed the case, I would still like to see how this goes wrong.