We know that the string order of a spin chain is defined as $$\mathcal{O}^\alpha=\lim_{i-j\to\infty}\left\langle S_i^\alpha\prod_{k=i+1}^{j-1}\exp(i\pi S_k^\alpha)\ S_j^\alpha \right\rangle$$ now consider a spin-1 chain and the state to be $|+1,+1,\cdots,+1\rangle$. Then the string order should be $\langle1\cdot(-1)^{j-i-1}\cdot1\rangle$, then this number is not defined.
I think the way to avoid this is by consider that any state cannot be so pure so that as $i-j\to\infty$ this number could have exponential decay or something like that, making the string order turns into 0. I'm highly likely to be wrong, but what the correct way to explain this?
Edit: as suggested, the string order might only be able to study the ground state topological properties, and for systems (Hamiltonian) with different class of symmetry, the string order parameter is different. However, what if one want to study an eigenstate's topological properties in a non-equilibrium scenario?