What you call "regular" measurement is probably projective measurement onto the computational basis, $|0\rangle$ and $|1\rangle$. Most introductory quantum computing resources define a qbit as follows:
$|\psi\rangle = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \alpha|0\rangle + \beta|1\rangle$
Where $|\alpha|^2$ gives you the probability of collapse to $|0\rangle$, and $|\beta|^2$ gives you the probability of collapse to $|1\rangle$. If this is what you know as "regular" measurement, then yes it's just projective measurement.
In addition to measuring in the computational basis, you can do a projective measurement onto any pair of orthogonal unit vectors. You do this by rewriting your quantum state in that basis. Here is a good video tutorial on how to do change-of-basis. For example, we can measure in the $X$ basis:
$|+\rangle = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}$, $|-\rangle = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} \end{bmatrix}$
Consider the following qbit value:
$|\phi\rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix} = 1|0\rangle + 0|1\rangle$
This qbit value has a 100% chance of collapsing to $|0\rangle$ when measured in the computational basis. What if we were to measure it in the $X$ basis? First we have to do a change-of-basis to the $X$ basis, where it is written as:
$|\phi\rangle = \frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}}|-\rangle$
Note we did not change the value of the qbit at all, we are just writing it a different way (in a different basis). Now we see that if we measure $|\phi\rangle$ in the $X$ basis, it has a 50% chance of collapsing to $|+\rangle$ and a 50% chance of collapsing to $|-\rangle$.
It's called projective measurement because geometrically you're "projecting" your quantum state vector onto the measurement basis vectors, and the length of the projection on a basis vector gives you the probability of collapsing to that basis vector.
