Assume $p$ has a $\text{Uniform}([0,\theta])$ prior, where $\theta\in(0,1)$ is a known constant. Let $X\sim \text{Binomial}(n,p)$ for some $n\in\mathbb{N}$. What is posterior $\mu(p|X)$?
(In the $\theta=1$ case it's well known that $p|X=x\sim\text{Beta}(1+x,1+n-x)$. What happens when $\theta<1$?)
Edit: Assume for simplicity that this prior is correctly specified. (i.e. $p$ is "actually" drawn according to $\text{Uniform}([0,\theta])$, and then $X$ is drawn as above given $p$). However, if you know the answer for the case where this prior is misspecified (prior is $\text{Uniform}([0,\theta])$, but $p$ is actually drawn according to $\text{Uniform}([0,1])$) then even better!
Why not just do the calculation:
$$p \sim \operatorname{Uniform}(0,\theta), \quad f(p) = \frac{1}{\theta} \mathbb 1(0 \le p \le \theta),$$ and $$X \mid p \sim \operatorname{Binomial}(n, p),$$ so that $$f(p \mid x) \propto \Pr[X = x \mid p]f(p) = p^x (1-p)^{n-x} \mathbb 1(0 \le p \le \theta).$$ This is a truncated beta; to show this, we integrate with respect to $p$ to find the normalization constant that makes the posterior likelihood a proper density:
$$\int_{p=0}^\theta f(p \mid x) \, dp = \int_{p = 0}^\theta p^x (1-p)^{n-x} \, dp$$ which is an incomplete beta function. For integers satisfying $0 \le x \le n$, there is a closed polynomial form; e.g., if $n = 7$ and $x = 3$, we get $\frac{\theta^4}{280}(70-224\theta+280\theta^2-160\theta^3+35\theta^4)$, and then the posterior density is $p^3(1-p)^4$ divided by this amount. It is not difficult to write this polynomial in terms of a finite sum; e.g., $$\begin{align} \int_{p=0}^\theta p^x(1-p)^{n-x} \, dp &= \int_{p=0}^\theta p^x \sum_{k=0}^{n-x} \binom{n-x}{k} (-p)^k \, dp \\ &= \sum_{k=0}^{n-x} \binom{n-x}{k} (-1)^k \int_{p=0}^\theta p^{x+k} \, dp \\ &= \sum_{k=0}^{n-x} \binom{n-x}{k} (-1)^k \frac{\theta^{x+k+1}}{x+k+1}. \end{align}$$ So an explicit answer for your posterior density would be:
$$f(p \mid x) = \frac{p^x (1-p)^{n-x} \mathbb 1(0 \le p \le \theta)}{\sum_{k=0}^{n-x} \binom{n-x}{k} (-1)^k \frac{\theta^{x+k+1}}{x+k+1}}.$$