Rust 如何初始化match声明中的两个变量而不会激怒borrow 判断器

所以我有这个使用rusqlite的代码.而且效果很好.

pub struct Database {
    conn: Connection,
}

impl Database {
    pub fn get(self: &Self, id: Option<u64>, name: Option<&str>) -> Result<Option<Record>, Error> {

        let mut stmt = match name {
            Some(_) => { self.conn.prepare_cached(Self::STMT_GET_ID_NAME) },
            None => { self.conn.prepare_cached(Self::STMT_GET_ID) }
        }?;

        let mut rows = match name {
            Some(name) => { stmt.query(params![id, name]) },
            None => { stmt.query(params![id]) }
        }?;


        let row = rows.next()?;
        if let Some(row) = row {
            let record = Record { 
                id: row.get(0)?, 
                parent: row.get(1)?, 
                name: row.get(2)?, 
                record_type: row.get(3)?,
                timestamp: row.get(4)?, 
                created_at: row.get(5)?, 
                modified_at: row.get(6)?, 
                size: row.get(7)?, 
                hash: row.get(8)?, 
                inode: row.get(9)? 
            };
            Ok(Some(record))
        } else {
            Ok(None)
        }
    }
}

我遇到的问题是match陈述本质上是重复的.

我try 过这样的事情，但这行不通.

        let (mut stmt, mut rows) = match name {
            Some(name) => {
                let mut stmt = self.conn.prepare_cached(Self::STMT_GET_ID_NAME)?;
                let rows = stmt.query(params![id, name])?;
                (stmt, rows)
            }
            None => {
                let mut stmt = self.conn.prepare_cached(Self::STMT_GET_ID)?;
                let rows = stmt.query(params![id])?;
                (stmt, rows)
            }
        };

问题是编译器抱怨stmt的生命周期

error[E0597]: `stmt` does not live long enough
error[E0505]: cannot move out of `stmt` because it is borrowed

我真的try 了很多东西，并try 疯狂地搜索.但我被困住了.

there甚至是一种不完全丑陋和惯用的方法吗？我觉得这是一个非常愚蠢的问题，而且我在这里错过了一些基本的东西..