字符串连接错误:需要文字 [英] string concat error: expected a literal

问题描述

所以我有这个:

struct User {
    reference: String,
    email: String,
    firstname: String,
    lastname: String
}

fn main() {

    let user = User {
        reference: "ref".to_string(),
        email: "[e-mail]".to_string(),
        firstname: "John".to_string(),
        lastname: "Doe".to_string()
    };

    concat!(&user.firstname.as_string(), " ", &user.lastname.as_string());

}

返回错误:

error: expected a literal
concat!(&user.firstname.as_string(), " ", &user.lastname.as_string());
        ^~~~~~~~~~~~~~~~~~~~~~~~~~

但我认为 .as_string() 已经使它成为一个文字，不是吗?我还发现到处都是 as_slice() 和 as_str() 引用，这很令人困惑.是哪个?

But I thought that .as_string() already makes it a literal, does it not? I am also finding as_slice() and as_str() references all over the place and it's confusing. Which one is it?

更新好吧，我希望我不必在这里粘贴整个内容，但我想无论如何我都必须这样做:

UPDATE OK, I was hoping I don't have to paste the whole thing here, but I guess I just have to do that anyway:

extern crate postgres;

use postgres::{Connection, SslMode};

struct User {
    reference: String,
    email: String,
    firstname: String,
    lastname: String
}

fn main() {

    let conn = Connection::connect("postgres://postgres:postgres@localhost/mydb", &SslMode::None).unwrap();

    let user = User {
        reference: "ref".to_string(),
        email: "[e-mail]".to_string(),
        firstname: "John".to_string(),
        lastname: "Doe".to_string()
    };
    let query = "INSERT INTO foo (user_id, name) VALUES ((SELECT id FROM user WHERE email = $1), $2)";
    conn.execute(query, &[&user.email, concat!(&user.firstname.as_slice(), " ", &user.lastname.as_slice())]).unwrap();

}

推荐答案

这里有误会.

concat! 是一个宏，即代码生成"机制；因此它在编译的第一阶段被扩展，在类型解析/所有权检查/等之前......

concat! is a macro, that is a "code-generation" mechanism; as a result it is expanded at the first stages of the compilation, before type resolution/ownership checks/etc...

文字是代码中按原样写入的值:true, 1, "hello";表达式的结果不能是文字(根据定义).结果类型可能看起来相似(甚至完全相同)，但类型在这里无关.

A literal is a value written as-is in the code: true, 1, "hello"; the result of an expression cannot be a literal (by definition). The resulting types may look similar (or even be identical) but types are irrelevant here.

那么，你真正想要的是什么?我猜你只是想连接字符串.对于 String，你可以使用 +:

So, what do you really want? I guess you just want to concatenate strings. For String, you can just use +:

let fullname = user.firstname + " " + user.lastname;
conn.execute(query, &[&user.email, &fullname]).unwrap();

或者，如果您需要一些更复杂的格式，您可以使用 format! 宏(不需要文字)，这里是:

Alternatively, if you need some more complex formatting, you can use the format! macro (which does not require literals), here it would be:

let fullname = format!("{} {}", user.firstname, user.lastname);

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