Question

InsertEmployee中的最后一项陈述是FetchEmployeesAsync。但FetchEmployeesAsync有一个await。据我所知，如果等待的任务没有完成，控制权将转移回呼叫者。但是执行仍然会在InsertEmployee结束时等待，而不是去调用者。我是对的吗？

public void InsertEmployee(Employee e)
{
    SQLiteAsyncConnection conn = new SQLiteAsyncConnection("Employee.sqlite");
    conn.InsertAsync(e);
    FetchEmployeesAsync();
 }


public async void FetchEmployeesAsync()
{
    SQLiteAsyncConnection conn = new SQLiteAsyncConnection("Employee.sqlite");
    employees = await conn.Table<Employee>().ToListAsync();
    DisplayList();
}

Answer 1

由于conn.Table<Employee>等待InsertEmployee调用，它将在任务完成时将控制权返回给其调用者。由于此方案中的调用方为FetchEmployees，无需等待即调用InsertEmployee，因此InsertEmployee方法将立即完成并退出。

当在堆栈中进一步等待任务时，控制将一直返回到聚合根，除非在到达聚合根的路上某个时候等待它，这通常是（但不总是）用户界面。

如果您想在FetchEmployees完成之前阻止FetchEmployees方法完成，可以让public async void InsertEmployee(Employee e) { SQLiteAsyncConnection conn = new SQLiteAsyncConnection("Employee.sqlite"); conn.InsertAsync(e); await FetchEmployees(); } public async Task FetchEmployeesAsync() { SQLiteAsyncConnection conn = new SQLiteAsyncConnection("Employee.sqlite"); employees = await conn.Table<Employee>().ToListAsync(); DisplayList(); }返回任务。

FetchEmployeesAsync

虽然Task的返回值为<?php $url = 'http://www.dandwiki.com/wiki/SRD:Classes'; $html = file_get_contents($url); $dom = new DOMDocument(); //from file @$dom->loadHTML($html); //Creating a new DOMPath $Xpath = new DOMXPath($dom); // Get links // //big Selects all <big> elements in the document. // [text()="Base Classes"] with text "base classes" // /.. get parent class // /.. get parent class // //a select all <a> elements // /text() get text $query = '//big[text()="Base Classes"]/../..//a/text()'; $entries = $Xpath->query($query); foreach ($entries as $entry) { echo $entry->nodeValue . "<br>"; } ?>，但您不必明确返回一个。在编译过程中会发生一些神奇的事情，当等待发生时，会为您返回一个任务，其余的方法被封送到一个延续

执行Async

1 个答案: