7. Given the function $f(x)=x^{3}+\sin x(x \in \mathbf{R})$, the function $g(x)$ satisfies $g(x)+g(2-x)=0(x \in \mathbf{R})$, if the function $h(x)=f(x-1)-g(x)$ has exactly 2019 zeros, then the sum of all these zeros is
From $f(-x)=(-x)^{3}+\sin (-x)=-\left(x^{3}+\sin x\right)=-f(x)$, we know that $f(x)$ is an odd function, and its graph is centrally symmetric about the point $(0,0)$. Therefore, the graph of $f(x-1)$ is centrally symmetric about the point $(1,0)$.

From $g(x)+g(2-x)=0$, we know that $g(x+1)=-g(1-x)$, so the graph of $g(x)$ is centrally symmetric about the point $(1,0)$. Thus, the graph of $h(x)=f(x-1)-g(x)$ is also centrally symmetric about the point $(1,0)$.

Given that $h(x)$ has 2019 zeros, the center of symmetry $(1,0)$ is also a zero point. Therefore, the remaining 2018 zeros are paired and symmetric about the point $(1,0)$. Consequently, the sum of all these zeros is $2 \times 1009 + 1 = 2019$. Hence, the answer is 2019.