7. If the parabola $y=\frac{1}{2} x^{2}-m x+m-1$ intersects the $x$-axis at integer points, then the equation of the axis of symmetry of the parabola is $\qquad$.
7. $x=1$.

From the problem, we know that the equation $\frac{1}{2} x^{2}-m x+m-1=0$ has two integer roots $x_{1} 、 x_{2}\left(x_{1} \leqslant x_{2}\right)$. Therefore,
$$
\frac{1}{2} x^{2}-m x+m-1=\frac{1}{2}\left(x-x_{1}\right)\left(x-x_{2}\right) \text {. }
$$

Substituting $x=1$, we get $\left(1-x_{1}\right)\left(1-x_{2}\right)=-1$.
Thus, $x_{1}=0, x_{2}=2$.
Hence, the equation of the axis of symmetry of the parabola is $x=1$.