Is every Eulerian graph also Hamiltonian?

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I know that if a graph is Eulerian then there exists an Eulerian cycle that contains all edges of the graph. I also know that if a graph is Hamiltonian then there exists a Hamiltonian cycle that contains all vertices of the graph.

It is easy for me to observe that a Hamiltonian graph may not be Eulerian (because may exist edges not contained in the Hamiltonian cycle). However, I'm a bit confused about the other direction. Is all Eulerian graphs also Hamiltonian?

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It is not the case that every Eulerian graph is also Hamiltonian. It is required that a Hamiltonian cycle visits each vertex of the graph exactly once and that an Eulerian circuit traverses each edge exactly once without regard to how many times a given vertex is visited. Take as an example the following graph:

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It's easy to find an Eulerian circuit, but there is no Hamiltonian cycle because the center vertex is the only way one can get from the left triangle to the right.

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Counterexample: the complete bipartite graph $K_{2,4}$ is Eulerian but not Hamiltonian.