7. Equation
$$
\sin \pi x=\left[\frac{x}{2}-\left[\frac{x}{2}\right]+\frac{1}{2}\right]
$$

The sum of all real roots of the equation in the interval $[0,2 \pi]$ is $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
7.12. Let $\left\{\frac{x}{2}\right\}=\frac{x}{2}-\left[\frac{x}{2}\right]$. Then for any real number $x$, we have $0 \leqslant\left\{\frac{x}{2}\right\}<1$.
Thus, the original equation becomes
$$
\sin \pi x=\left[\left\{\frac{x}{2}\right\}+\frac{1}{2}\right] \text {. }
$$
(1) If $0 \leqslant\left\{\frac{x}{2}\right\}<\frac{1}{2}$, then
$$
\begin{array}{l}
\Rightarrow \sin \pi x=\left[\left\{\frac{x}{2}\right\}+\frac{1}{2}\right]=0 \\
\Rightarrow \pi x=k \pi(k \in \mathbf{Z}) \\
\Rightarrow x=k(k \in \mathbf{Z}) .
\end{array}
$$

Combining $x \in[0,2 \pi]$, we know $x=0,1, \cdots, 6$.
Upon inspection, $x=0,2,4,6$ meet the requirements of the problem.
$$
\begin{array}{l}
\text { (2) If } \frac{1}{2} \leqslant\left\{\frac{x}{2}\right\}<1 \\
\Rightarrow \sin \pi x=\left[\left\{\frac{x}{2}\right\}+\frac{1}{2}\right]=1 \\
\Rightarrow \pi x=2 k \pi+\frac{\pi}{2}(k \in \mathbf{Z}) \\
\Rightarrow x=2 k+\frac{1}{2}(k \in \mathbf{Z}) .
\end{array}
$$

In this case, $\left\{\frac{x}{2}\right\}=0$, which does not meet the requirements.
Therefore, the values of $x$ that satisfy the conditions are $0,2,4,6$, and their sum is 12.