1. Given the sets $P=\left\{x \mid x^{2}=1\right\}$ and $Q=\{x \mid m x=1\}$. If $Q \subset P$, then the number of possible values for the real number $m$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
$-1 .(\mathrm{D})$.
$$
\because P=\{-1,1\}, Q \subset P \text {, }
$$
$\therefore Q$ could be $\varnothing,\{-1\},\{1\}$.
When $Q=\varnothing$, $m=0$;
When $Q=\{-1\}$, $m=-1$;
When $Q=\{1\}$, $m=1$.
Thus, $m$ has 3 values: $0,-1,1$.