There is a formula that works in all degrees, not just imaginary quadratic. In a global field $K$, let $O$ be integral over ${mathbf Z}$ or ${mathbf F}[t]$ (${mathbf F}$ a finite field) and be "big", i.e., it has fraction field $K$. Let $mathfrak c$ be the conductor ideal of $O$ in its integral closure $R$. Then
$$h(O) = frac{h(R)}{[R^times:O^times]}frac{varphi_{R}({mathfrak c})}{varphi_O(mathfrak c)},$$
where $varphi_O(mathfrak c)$ is the number of units in $O/mathfrak c$ and $varphi_R(mathfrak c)$ is the number of units in $R/mathfrak c$. This is derived in Neukirch's alg. number theory book in the number field case, but it goes through to any one-dimensional Noetherian domain with a finite residue rings and a finite class group.
In the imag. quadratic case the unit index $[R^times:O^times]$ is 1 most of the time so you don't notice it.
Both $varphi_R(mathfrak c)$ and $varphi_O(mathfrak c)$ can be written in the form ${text N}(mathfrak c)prod_{mathfrak p supset mathfrak c}(1 - 1/{text{N}(mathfrak p)})$, where the ideal norm $text N$ means the index in $R$ or $O$ and $mathfrak p$ runs over primes in $R$ or $O$ for the two cases.
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