Is the integral closure of a local domain in a finite extension of its field of fractions semi-local? 
Is the integral closure of a local domain in a finite extension of its field of fractions semi-local?

If the answer is negative, I wonder under what conditions it would be semi-local.
EDIT
Here's an example of a local domain which is not necessarily a Japanese ring.
Let $A$ be a valuation ring, $K$ its field of fractions.
Let $P$ be the maximal ideal of $A$.
Let $L$ be a finite extension of $K$.
Let $B$ be the integral closure of $A$ in $L$.
It is well-known that there exist only finitely many valuation rings of $L$ dominating $A$.
Let $M$ be a maximal ideal of $B$.
It is well-known that $M$ lies over $P$.
Hence $B_M$ dominates $A$.
There exists a valuation ring $R$ of $L$ dominating $B_M$.
Since $M$ is determined by $R$ and $R$ dominates $A$, there exist only finitely many maximal ideals of $B$.
Hence $B$ is a semilocal ring.
Note that $B$ is not necessarily a finite $A$-module(even if $A$ is a discrete valuation ring).
Hence $A$ is not necessarily a Japanese ring.

Here is a non-Noetherian result [Bourbaki, Commutative algebra, VI, $\S$8, no.3, Remarque]: For a valuation ring $R$ with fraction field $K$ and a finite field extension $L/K$, the integral closure of $R$ in $L$ is semilocal.
