Let $\{(N_j, B_j, L_j): 1 \le j \le m\}$ be finitely many Hadamard triples in $\mathbb{R}$. Given a sequence of positive integers $\{n_k\}_{k=1}^\infty$ and $\omega=(\omega_k)_{k=1}^\infty \in \{1,2,\cdots, m\}^\mathbb{N}$, let $\mu_{\omega,\{n_k\}}$ be the infinite convolution given by $$\mu_{\omega,\{n_k\}} = \delta_{N_{\omega_1}^{-n_1} B_{\omega_1}} * \delta_{N_{\omega_1}^{-n_1} N_{\omega_2}^{-n_2} B_{\omega_2}} * \cdots * \delta_{N_{\omega_1}^{-n_1} N_{\omega_2}^{-n_2} \cdots N_{\omega_k}^{-n_k} B_{\omega_k} }* \cdots. $$ In order to study the spectrality of $\mu_{\omega,\{ n_k\}}$, we first show the spectrality of general infinite convolutions generated by Hadamard triples under the equi-positivity condition. Then by using the integral periodic zero set of Fourier transform we show that if $\mathrm{gcd}(B_j - B_j)=1$ for $1 \le j \le m$, then all infinite convolutions $\mu_{\omega,\{n_k\}}$ are spectral measures. This implies that we may find a subset $\Lambda_{\omega,\{n_k\}}\subseteq \mathbb{R}$ such that $\big\{ e_\lambda(x) = e^{2\pi i \lambda x}: \lambda \in \Lambda_{\omega,\{n_k\}} \big\}$ forms an orthonormal basis for $L^2(\mu_{\omega,\{ n_k\}})$.
Comment: 21 pages; final version