Computing degrees of projective varieties via Chern classes I know that the degree of a projective hypersurface $H \subset \mathbb{P}^n$ can be computed in terms of the Chern class of the normal (line) bundle of $H$. Is there a similar formula for the degree of a higher codimension projective variety in terms of Chern classes of the normal bundle? 
In general, does degree just depend on the normal bundle of the projective variety in projective space? I feel like the answer is no, which would make it impossible to compute the degree in terms of the Chern class of the normal bundle. 

Here is a general answer in terms of Chow rings.  
Let $i:Y\hookrightarrow \mathbb P^n$ be a smooth closed subvariety of codimension $r$ and degree $d$, so that for the corresponding cycle class we have  $[Y]=dH^r\in A^r(\mathbb P^n)=\mathbb Z\cdot  H^r$ .
We have (Hartshorne, page431): $$c_r(N_{Y/X})=i^\ast [Y]=i^\ast (dH^r)=dh^r\in A^r(Y)$$ where $h=i^\ast H\in A^r(Y)$ and where $c_r(N_{Y/X})\in A^r(Y)$ is the Chow Chern class
 [which, in case the base field is $\mathbb  C$, is infinitely more precise than its image in singular cohomology $c_r^\mathbb C(N_{Y/X})\in H^{2r}(Y(\mathbb C),\mathbb Z)$ ].
If $c_r(N_{Y/X})$ is known, we may often extract the degree $d$ of $Y$ from the the above formula $c_r(N_{Y/X})=dh^r$.
 However if  $h^r=0$, as is the case for curves in $\mathbb P^3$ for example, the equality $c_r(N_{Y/X})=dh^r$ reduces to $0=d\cdot0$, which (in conformity with Pooh Bear's great counterexample) doesn't allow us to compute $d$. 
