替换修饰符中的参数扩展

Question

Let x be a parameter initialized as follows:

x=("drink/beer" "drink/whine" "drink/boooze")

Suppose there are also parameters y and z , with y="drink" amd z="buy" , respectively. For each word in x , I want to substitute the portion that machtes y with z . Expected output , hence, is

buy/beer buy/whine buy/boooze

I though the following code would do so:

print ${x:s/${y}/${z}}

However, it does not work. The "drink" substrings are not substituted and I could find the reason in themodifier documentation :

The s/l/r/ substitution works as follows [...] When used in a history expansion, which occurs before any other expansions, l and r are treated as literal strings

This means, that in my attempt zsh literally searchs for "${y}", which is of course not to be found. Is there another way the accomplish my goal, without using an explicite loop?

Answer 1

Parameter expansion has its own replacement syntax, separate from the history modification operators.

% print ${x/$y/$z}
buy/beer buy/whine buy/boooze