.net 等待时 Blazor 服务器按钮刷新

您的代码将只工作IF do some workyield 率.如果它是包装在Task中的同步代码块，那么在do some work完成之前，呈现器没有线程时间来更新UI，并且您已经将状态字段设置回已完成.

引述今天另一项类似问题的答案:

调用StateHasChanged不会呈现组件.它将组件的RenderFragment放在呈现器的队列中.只有当呈现器的线程时间运行时，该队列才会[由呈现器和所呈现的组件]得到服务.

在下面的代码中，我引入了状态字段更新后的Year，以确保在该点上更新UI.

@page "/"

<PageTitle>Index</PageTitle>

<h1>Hello, world!</h1>

<div class="m-2 p-2">
    <button type="button" class="btn btn-primary" @onclick="DoSearch" disabled="@SearchDisabled">@this.SearchBtnName</button>
</div>

@code {
    protected bool SearchDisabled = false;
    protected string SearchBtnName = "Search";

    private async Task DoSearch()
    {
        SearchDisabled = true;
        SearchBtnName = "Searching ...";

        // make sure we yield to update the UI
        // Do work may be a block of synchronous wrapped in a Task!
        // Some people will suggest you do await Task.Delay(1) here Take your choice
        await Task.Yield();
        //await Task.Delay(1);

        await this.DoWork();

        SearchDisabled = false;
        SearchBtnName = "Search";
    }

    private async Task DoWork()
    {
        // emulate real async work
        await Task.Delay(2000);
    }
}