I feel really bad for the OP getting these downvotes. The question (if worded better) is a really good question. There is obviously a lot of confusion over this - given the range of solutions posted. I think the main confusion comes from two places: (1) using current density in place of current (whether consciously or not), (2) Not realizing that a 1-D vector is a scalar.
Current density:
For simplicity, assume we have a conductor with a constant conductance, $\sigma$. Also for simplicity, let's assume that the electric field is time-independent, but could vary in space. The electric field, $\vec{E}(\vec{x})$, is a vector field, that is, at each point in space, we associate a vector space. 359
The relationship between the electric field and the current density in this simplified model is $$\vec{J}(\vec{x}) = \sigma~\vec{E}(\vec{x})$$ From this equation, we see that $\vec{J}(\vec{x})$, the current density, is also a vector field.
Current:
The current is defined with reference to some area as $$I_A = \int_A \vec{J}(\vec{x})\cdot d\vec{A}$$ In circuits, the area we care about is the cross-sectional area of the wire. With a DC source the electric field is constant throughout the wire (if AC, still constant spatially but has time dependence), that is $\vec{E}(\vec{x}) = E_0~\hat{z}$ (assuming the wire is along the $z$-axis). This means that the current density is $$\vec{J} = \sigma~E_0~\hat{z}$$ and so the current is (area of interest is the cross-section of the wire, with $d\vec{A} = dA\hat{z}$) $$I = \int dA~\sigma E_0~\hat{z}\cdot\hat{z} = A\sigma ~ E_0$$ In this idealized situation, we see that $I$ can be treated as a 1-D vector, so far as the vector space axioms are concerned. We also note that a 1-D vector is a scalar! Note that I am not saying scalar field or vector field.
Let's see what happens when we aren't confined to a wire and have spatial varying electric fields.
Take some area through a region in a conductor (with constant $\sigma$ as before) where the electric field has spatial dependence $\vec{E}(\vec{x})$. The current associated with this electric field and area is $$I_A = \sigma\int_A \vec{E}(\vec{x})\cdot d\vec{A}$$ With a little thought, one can see that this doesn't change the argument in our wire case. The current defined this way still produces a 1-D vector, whose direction indicates which way the total flow of positive charges (by convention) is flowing, through the prescribed surface. And again, a 1-D vector IS a scalar.
Note: in this situation, adding currents is physically identical to adding electric fields (as $\sigma$ is assumed constant). 780