如何在 zsh 中获取 bash 脚本

Question

I have bash script which heavily uses bash features and does not run in zsh, dash, and sh.

My prefered shell is zsh. How can I source this bash script into zsh shell?

I tried

function in() {
    bash -c "'$@ ; exec zsh'"
}

but this gives me syntax errors like this

❯ in source t.bash
t.bash ; exec zsh': -c: line 1: unexpected EOF while looking for matching `''
t.bash ; exec zsh': -c: line 2: syntax error: unexpected end of file

I also tried to use xargs but then enviroment variables set by my bash script are not set in zsh

function in () {
   echo -c "'$@ ; exec zsh'" | xargs -t bash
}

but then enviroment variables set by my bash script are not set in zsh

copy pasting the command shown by xargs -t does work

Answer 1

You said "copy pasting the command shown by xargs -t does work".

So... if you changed

function in () {
   echo -c "'$@ ; exec zsh'" | xargs -t bash
}

to

function in () {
   echo -c "'$@ ; exec zsh'" | xargs -t bash >"${BATCH_FILE}"
   chmod 754 "${BATCH_FILE}" ; "${BATCH_FILE}"
}

wouldn't that give you what you need?