Question

这是我的代码

extern crate postgres;

use postgres::{Connection, SslMode};

struct User {
    reference: String,
    email: String
}

static DB_URI: &'static str = "postgres://postgres:postgres@localhost/test";

fn main() {

    let conn = Connection::connect(DB_URI, &SslMode::None).unwrap();
    let trans = conn.transaction().unwrap();

    let user = User {
        reference: "123abc".to_string(),
        email: "test@test.com".to_string()
    };

    let result = insert_user(&trans, &user);

    trans.set_commit();
    trans.finish();

}

fn insert_user<'_>(trans: &postgres::Transaction<'_>, user: &User) -> postgres::Result<postgres::rows::Rows<'_>> {
    let query = "INSERT INTO usr (reference, email) VALUES ($1, $2)";
    trans.prepare(query).unwrap().query(&[&user.reference, &user.email])
}

它产生了一个错误：

src/main.rs:31:2: 31:31 error: borrowed value does not live long enough
src/main.rs:31  trans.prepare(query).unwrap().query(&[&user.reference, &user.email])
                ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~
src/main.rs:29:114: 32:2 note: reference must be valid for the lifetime '_ as defined on the block at 29:113...
src/main.rs:29 fn insert_user<'_>(trans: &postgres::Transaction<'_>, user: &User) -> postgres::Result<postgres::rows::Rows<'_>> {
src/main.rs:30  let query = "INSERT INTO usr (reference, email) VALUES ($1, $2)";
src/main.rs:31  trans.prepare(query).unwrap().query(&[&user.reference, &user.email])
src/main.rs:32 }
src/main.rs:29:114: 32:2 note: ...but borrowed value is only valid for the block at 29:113
src/main.rs:29 fn insert_user<'_>(trans: &postgres::Transaction<'_>, user: &User) -> postgres::Result<postgres::rows::Rows<'_>> {
src/main.rs:30  let query = "INSERT INTO usr (reference, email) VALUES ($1, $2)";
src/main.rs:31  trans.prepare(query).unwrap().query(&[&user.reference, &user.email])
src/main.rs:32 }
error: aborting due to previous error
Could not compile `test`.

这里有什么问题？

Answer 1

我认为问题在于：

fn insert_user<'a>(trans: &postgres::Transaction<'a>, user: &User) -> postgres::Result<postgres::rows::Rows<'a>> {

（我已将生命周期参数名称更改为通常的名称）

在这里，您要说明结果中Rows参数的生命周期应与Transaction参数中的生命周期相同（基本上是Connection对象的生命周期）。但是，Rows的生命周期参数等于Statement的生命周期，Statement值（通过调用prepare()方法创建）是局部变量，因此它是严格小于要求（局部变量的寿命总是小于参数中指定的寿命）。

此错误是合法的 - Rust在此处阻止了实际的逻辑错误。 Rows迭代器需要Statement来加载其数据，但在这种情况下，Statement会在Rows仍然存活时被销毁。

您需要做的是从Rows收集数据到某个容器（例如Vec）并将其返回。但是，insert_user()似乎是一个不从数据库返回任何内容的查询。对于此类查询，您应该在Statement上使用execute()方法，并且您的函数应如下所示：

fn insert_user(trans: &postgres::Transaction, user: &User) -> postgres::Result<u64> {
    let query = "INSERT INTO usr (reference, email) VALUES ($1, $2)";
    trans.prepare(query).unwrap().execute(&[&user.reference, &user.email])
}

借来的价值不够长

1 个答案: