I suspect the following works. Let $\mathcal{X}$ be a proper, smooth, finite type Deligne-Mumford stack over $k$ that is one-dimensional and that has a dense open substack $U$ that is a scheme. Let $u:\mathcal{X}\to X$ be the coarse moduli space. Let $\{p_1,\dots,p_r\}\subset X$ be the complement of $U$. For each point $p_i$ of $D$, let the order of the corresponding stabilizer be $n_i$. Let $n$ be the least common multiple over every integer $n_i$. Define $m_i = n/n_i$.

Consider the divisor $D_0$ on $X$ that is $$m_1\underline{p}_1 + \dots m_r\underline{p}_r + n\underline{q}_1+ \dots + n\underline{q}_s,$$
where $q_1,\dots,q_s$ are points different from $p_1,\dots,p_r$. For $s$ sufficiently large, $D_0$ is linearly equivalent to an effective Cartier divisor $D_\infty$ that is disjoint from $D_0$. These two divisors together define a morphism $f:X\to \mathbb{P}^1$ such that the preimage of $\{0\}$ is $D_0$ and the preimage of $\{\infty\}$ is $D_\infty$.

Now let $v:\mathbb{P}(1,n)\to \mathbb{P}^1$ be the coarse moduli space, where the unique stacky point of $\mathbb{P}(1,n)$ maps to $0$. Since $u^{-1}f^{-1}(\{0\})$ is divisible by $n$ as an effective divisor on $\mathcal{X}$, it seems to me that $f\circ u$ factors through $v$, i.e., there is a unique $1$-morphism $F:\mathcal{X}\to \mathbb{P}(1,n)$ such that $v\circ F$ is $2$-equivalent to $f\circ u$.

Since $f$ is finite, the only thing to check is that $F$ is representable. Of course $F$ is representable away from the stacky points $p_i$. Since $n_i$ divides $n$, also $F$ should be representable at $p_i$.