RUST - 返回任务的组合结果

Question

情况

我有以下方法：

pub async fn get_projects(organizations: &Vec<Organization>) -> Vec<Project> {
    // Define collection to hold the projects.
    let mut result = Vec::new();
    // Create a collection of tasks to retrieve the projects.
    let mut tasks = Vec::new();
    // Create a new task for each organization.
    for organization in organizations.iter() {
        if !organization.is_pat_valid {
            // If the PAT for the organization is not valid, do not try to retrieve the projects for it.
            continue;
        }
        // Create a new task to retrieve the projects for the organization and store it.
        tasks.push(repositories::devops::get_projects(
            &organization.pat,
            &organization.name,
        ))
    }
    // Wait for all tasks to finish.
    let task_results = join_all(tasks).await;
    // Add the projects from the resolved tasks all to the result collection.
    for task_result in task_results.iter() {
        match task_result {
            Ok(projects) => result.extend(projects),
            Err(error) => println!("Error: {}", error),
        }
    }    
    return result;
}

我想在这里做的是并行执行异步代码，最后等待所有任务，然后将所有任务结果一起添加到一个集合中，我返回。

问题

在代码部分：

for task_result in task_results.iter() {
        match task_result {
            Ok(projects) => result.extend(projects),
            Err(error) => println!("Error: {}", error),
        }
    }

我想添加到我的结果中的

projects

来自&Project类型，而不是Project。如果我将

result

vec 更改为

Vec<&Project>

我收到以下错误：

return result;
   |   ^^^^^^ returns a value referencing data owned by the current function

问题

我对 RUST 完全陌生，我不知道如何返回这个结果，因为似乎所有任务的结果都在我的 fn 范围内。

更多信息

旧的、性能不佳的解决方案是这样实现的：

pub async fn get_projects(organizations: Vec<Organization>) -> Vec<Project> {
    let mut result: Vec<Project> = vec![];
    for organization in organizations.iter() {
        if (!organization.is_pat_valid) {
            continue;
        }
        let organization_projects =
            repositories::devops::get_projects(&organization.pat, &organization.name).await;
        match organization_projects {
            Ok(projects) => {
                result.extend(projects);
            }
            Err(error) => {
                println!("Error: {}", error);
            }
        }
    }
    return result;
}

这应该更慢，因为 asyn fn 不是并行执行的，但至少它起作用了。