Does $\mathbb R \overset{\iota}{\hookrightarrow} \mathfrak{aff}(1,\mathbb R) \overset{\pi}{\twoheadrightarrow} \mathbb R$ Lie algebra extension exist?

Question

I think I have an example for a $\mathbb R \overset{\iota}{\hookrightarrow} \mathfrak{aff}(1,\mathbb R) \overset{\pi}{\twoheadrightarrow} \mathbb R$ Lie algebra extension. $$\iota:\mathbb R\to \mathfrak{aff}(1,\mathbb R): x\mapsto \begin{pmatrix}0 & x\\0 & 0\end{pmatrix}$$ is an embedding of $\mathbb R$ into $\mathfrak{aff}(1,\mathbb R)$ as an ideal, and the map $$\pi:\mathfrak{aff}(1,\mathbb R)\to\mathbb R: \begin{pmatrix}a & b\\0 & 0\end{pmatrix}\mapsto a$$ is a surjective projection onto $\mathbb R$ having kernel $\iota(\mathbb R)$, so with these functions, the sequence in question is exact, hence it is a Lie algebra extension. But I think such an extension should not exist because $H^2(\mathbb R,\mathbb R)=0$. Where am I wrong?