This is true. Consider the following lemma:
Let $X$ be an irreducible variety, $f\in k[X]$ a nonzero nonconstant function, and $Y$ an irreducible component of $V(f)$. $Y$ has codimension 1 in $X$.
Proof: We may restate as follows. Let $A$ be an an affine $k$-algebra which is a domain with $f\in A$ a nonzero nonunit. Let $P$ be a minimal prime ideal containin $f$- then $P$ must have height one. The height is at most one, by Krull's principal ideal theorem, and at least one, as $X$ irreducible.
I may then apply this lemma $t$ times to the chain $V(0)\supset V(P_1)\supset V(P_1,P_2)\supset\cdots\supset V(P_1,\cdots,P_t)$. This proves the claim. (The reason it must be at most instead of exactly is that a function $P_i$ may fail to be a nonzero nonunit on some irreducible component of $V(P_1,\cdots,P_j)$ where $i<j$. In this case, the final line of the lemma where it is shown that the height is at least one fails.)