What does it mean for a prime ideal to lie over another ideal in the case of a morphism of rings?

Question

When we usually talk about prime ideals "lying over" one another, we have a tower of rings $A \subseteq B$ with $\mathfrak{p} \subseteq A$ and $\mathfrak{q} \subseteq B$ prime ideals of $A$ and $B$ respectively. Then we can say that $\mathfrak{q}$ lies over $\mathfrak{p}$ if $$ A \cap \mathfrak{q} = \mathfrak{p}. $$ This is standard and I am comfortable with that. The issue is when texts have a morphism of rings $\phi: A \longrightarrow B$ and talk about primes of $B$ lying over primes of $A$. What exactly does this mean? The confusion stems from the fact that the extension of a prime ideal in $A$ is not generally a prime ideal in $B$, so there doesn't seem to be an obvious interpretation to me. In the case that $\phi$ is surjective, then it makes sense for primes containing the kernel of $\phi$, but in general, how does one interpret the notion of a prime of $B$ lying over a prime ideal when all we have is a general morphism of rings $\phi: A \longrightarrow B$?