Does every Noetherian domain have finitely many height 1 prime ideals?

Let $A$ be a Noetherian domain. Is the set $\{P\subset A \mid P \mbox{ prime ideal, } \dim A_P=1\}$ always finite?

I can prove for $f \neq 0, f\in A$, the set $\{P\subset A \mid \dim A_P=1, f\in P\}$ is finite (by using the primary decomposition of $\sqrt{(f)}$). The above statement is just the case when $f=0$.

No. Look at the ring $A=\mathbb{Z}$.

If $\dim A=0$, then the answer is obviouly yes.

If $\dim A=1$, then the answer is yes if and only if $A$ is semi-local (i.e. has only finitely many maximal ideals). This is because height one prime ideals are then maximal.

If $\dim A>1$, then there are always infinitely many prime ideals of height 1. Actually, $A$ has a prime ideal $\mathfrak m$ of height $>1$. EDIT [After taking the quotient of $A$ by a minimal prime ideal appearing in a chain of prime ideals contained in $\mathfrak m$ of positive length, one can assume that $A$ is integral.]End of Edit. Let $a\in \mathfrak m$ be non-zero. Then the minimal prime ideals over $a$ have height $1$ by Krull's principal ideal theorem. Pick one of them $\mathfrak p_1$. Pick $a_1\in \mathfrak m\setminus \mathfrak p_1$ and let $\mathfrak p_2$ be a minimal prime ideal over $a_1$. Then $\mathfrak p_2\ne \mathfrak p_1$. Let $a_2\in \mathfrak m\setminus (\mathfrak p_1\cup \mathfrak p_2)$ and pick a minimal prime ideal $\mathfrak p_3$ over $a_2$ and so on. You get an infinite sequence of prime ideals of height $1$.