Electric current has a magnitude given by $$\left|\int_\Sigma \mathbf{J}\cdot\mathbf{N}_e\,\mathrm d\sigma\right|$$where, in the surface integral, $\Sigma$ is the orientable surface through which the current flows, $\mathbf{J}$ is the current density and $\mathbf{N}_e$ is the external normal unit vector to $\Sigma$. You see that, by changing the orientation of the surface, the normal vector changes its sign and therefore the integral$$\int_\Sigma \mathbf{J}\cdot\mathbf{N}_e\,\mathrm d\sigma$$also changes its sign.
You can arbitrarily choose one of the two sides, having external normal $\mathbf{N}_{e,1}$, of the orientable surface $\Sigma$ in a way such that, at a particular instant, the current is $i\ge 0$ when $$\int_\Sigma \mathbf{J}\cdot\mathbf{N}_{e,1}\,\mathrm d\sigma\ge 0.$$Of course, if we call $\mathbf{N}_{e,2}$ the external normal on the other side of $\Sigma$, we have that $$\int_\Sigma \mathbf{J}\cdot\mathbf{N}_{e,1}\,\mathrm d\sigma\ge 0\iff \int_\Sigma \mathbf{J}\cdot\mathbf{N}_{e,2}\,\mathrm d\sigma\le 0.$$ In that sense, the direction of the current is represented by the orientation of the surface, considered to calculate it, that makes the integral $\int_\Sigma \mathbf{J}\cdot\mathbf{N}_{e,1}\,\mathrm d\sigma\ge 0$ positive.
So you have that current is not at all a vector belonging to $\mathbb{R}^3$ - while the current density $\mathbf{J}$, of course, is - because its direction can only be mono-dimensional: in or out of a side of the surface. If you wish, any real number can be seen as an element of a mono-dimensional vector space, in fact a structure of vector space is quite trivially defined on $\mathbb{R}$, but that is not, as far as I know, what is usually meant by vector quantity in elementary physics, which usually is an element of a vector of a space of dimension greater than $1$.