rust - 为自定义类型实现 AsExpression 和 FromSqlRow

Question

我在用：

diesel = { version = "1.3.0", features = ["postgres", "chrono"] }

该结构Payment包含一个自定义类型CentAmount。为了使包含结构可插入，我实现了

• AsExpression
• FromSqlRow

如简化自定义类型的支持中所述：

use diesel::expression::AsExpression;
use diesel::helper_types::AsExprOf;
use diesel::pg::Pg;
use diesel::row::Row;
use diesel::sql_types::{Integer, Nullable};
use diesel::types::FromSqlRow;
use std::error::Error;

#[derive(Serialize, Deserialize, Debug, PartialEq, Clone, Copy)]
pub struct CentAmount {
    cents: i32,
}

impl CentAmount {
    pub fn new(cents: i32) -> Self {
        assert!(cents >= 0);
        CentAmount { cents: cents }
    }
}

impl<'a> AsExpression<Nullable<Integer>> for &'a CentAmount {
    type Expression = AsExprOf<i32, Nullable<Integer>>;

    fn as_expression(self) -> Self::Expression {
        AsExpression::<Nullable<Integer>>::as_expression(self.cents)
    }
}

impl FromSqlRow<Integer, Pg> for CentAmount {
    fn build_from_row<R: Row<Pg>>(row: &mut R) -> Result<Self, Box<Error + Send + Sync>> {
        let cents = i32::build_from_row(row)?;
        Ok(CentAmount { cents })
    }
}

结构体在这里使用

#[derive(Queryable, Debug, Getters, Serialize, Deserialize, PartialEq, Insertable)]
#[table_name = "payments"]
pub struct Payment {
    pub id: Option<i32>,
    amount: CentAmount,
    leistung_id: i32,
    beschaeftiger_id: i32,
    payment_state: PaymentState,
    approval_date: Option<DateTime<Local>>,
}

我仍然收到错误：

the trait `diesel::Expression` is not implemented for `model::cent_amount::CentAmount`

我错过了什么？

标签： rustrust-diesel