Is charge 'localization' implicit in the idea of current? If it was possible for charge to assume arbitrary densities, like we often see electrostatic exercises, and one could spread charge density uniformly over a ring, then how one would, theoretically, distinguish between the situations when there is a current on the ring and when there is not? In both situations, the distributions of charges would be constant over the ring during time.
 A: I'm not sure what your question has to do with the title, but uniform charge density does not mean that the current is zero any more than uniform fluid density means that the velocity field of the fluid is zero.
For example, in quantum mechanics, a wavefunction may in some region take on the form $\psi(x) = A e^{i (k x - \omega t)}$, and the probability density is everywhere constant, $\rho = |A|^2$, whereas the probability current $j = \frac{\hbar}{2mi} \left( \psi^*\partial_x \psi - \partial_x \psi^* \psi \right) = \rho \frac{\hbar k}{m} = \rho v \neq 0$.
A: You can't find out the current by picking a time $t_1$ and paying people to tell you where all the continuous charge density is at at time $t_1$ and then picking another time $t_2$ and paying people to tell you where all the continuous charge density is at at time $t_2$.  As other answers mentioned you couldn't find out how fast water is flowing by doing the similar questions for a continuous water density.
One classical physics approach would be to track each ion or electron or other quanta of charge.  But in reality trying to measure it precisely will in general disturb the results about the current, so it is not a way to find the current.
The real way to find the current is with the magnetic field.  If you had a uniform charge distribution nailed to a disc you'd see an electrostatic field.  If you had the uniform charge nailed to a rotating disc, then you will also see a magnetic field.  You can also hollow out a little region and see the charge build up, another sign of current, different faces in different directions measure the components.
If you want to model it rather than measure it, then for each species of continuous charge with volume charge density $\rho_i(x,y,z)$ assign a vector field $\vec{v}_i(x,y,z)$ and then you get a volume current density $\vec{J}_i(x,y,z)=\rho_i(x,y,z)\vec{v}_i(x,y,z)$, do that for each charge density that needs to be moving at different velocities and add it up to get the total $\vec{J}=\vec{J}_1+\vec{J}_2+ ... +\vec{J}_k$, and that's the $\vec{J}$ that appears in: $$\vec{\nabla}\times \vec{B}=\mu_0(\vec{J}+\epsilon_0\frac{\partial \vec{E}}{\partial t}).$$
