将i32转换为&str时如何克服“借用时临时价值下降” [重复]

Question

这个问题在这里已经有了答案：

• 如何将临时字符串转换为 &str？ [重复] 1 个回答
• 如何制作格式？ 从条件表达式返回 &str？ 2 个答案
• 将本地字符串作为切片返回 (&str) 3 个答案

pub fn uidl(&mut self, message_number: Option<i32>) -> POP3Result {
    let arg = match message_number {
        Some(i) => Some(i.to_string().as_str()),
        None => None,
    };

    let command = match message_number {
        Some(_) => POP3Command::UidlOne,
        None => POP3Command::UidlAll,
    };

    self.execute_command(command, arg)
}

error[E0716]: temporary value dropped while borrowed
   --> src/pop3_stream.rs:156:29
    |
155 |         let arg = match message_number {
    |             --- borrow later stored here
156 |             Some(i) => Some(i.to_string().as_str()),
    |                             ^^^^^^^^^^^^^         - temporary value is freed at the end of this statement
    |                             |
    |                             creates a temporary which is freed while still in use
    |
    = note: consider using a `let` binding to create a longer lived value

如何摆脱这个错误？