我最近开始学习 Rust,我不确定如何从应该返回 Result 的函数返回未来值。当我尝试仅返回响应变量并删除结果输出时,出现错误:无法在返回的函数中使用运算符?std::string::String
#[tokio::main]
async fn download() -> Result<(),reqwest::Error> {
let url = "https://query1.finance.yahoo.com/v8/finance/chart/TSLA";
let response = reqwest::get(url)
.await?
.text()
.await?;
Ok(response)
}
Run Code Online (Sandbox Code Playgroud)
我在 main() 中期望的是获取并打印响应值:
fn main() {
let response = download();
println!("{:?}", response)
}
Run Code Online (Sandbox Code Playgroud)
我想你的代码应该是这样的
extern crate tokio; // 0.2.13
async fn download() -> Result<String, reqwest::Error> {
let url = "https://query1.finance.yahoo.com/v8/finance/chart/TSLA";
reqwest::get(url).await?.text().await
}
#[tokio::main]
async fn main() {
let response = download().await;
println!("{:?}", response)
}
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
8484 次 |
| 最近记录: |