Question

试图依次显示多条Toast消息，但仅显示最后一条Toast消息。我尝试使用Thread.sleep和处理程序来缓冲消息，但是都没有用。还有其他提示吗？这是我的代码：

public class GenericExceptionHandlerFilter implements Filter {
  private final List<HttpMessageConverter<?>> converters;

  public GenericExceptionHandlerFilter(<HttpMessageConverter<?>> converters) {
    this.converters = converters;
  }

  @Override
  public void doFilter(ServletRequest servletRequest, ServletResponse servletResponse,
      FilterChain filterChain) throws IOException {
    try {
      filterChain.doFilter(servletRequest, servletResponse);
    } catch (Exception e) {
      MyCommonResponse commonResponse = new MyCommonResponse(e);
      write(servletRequest, servletResponse, comonResponse);
      log.error("Unhandled exception.", e);
    }
  }

  private void write(HttpServletRequest servletRequest, HttpServletResponse servletResponse, Object object) throws IOException {
    String accept = servletRequest.getHeader("accept");
    for (HttpMessageConverter messageConverter : converters) {
      if (messageConverter.canWrite(object.getClass(), MediaType.valueOf(accept))) {
        HttpOutputMessage outputMessage = new ServletServerHttpResponse(servletResponse);
        messageConverter.write(object, MediaType.valueOf(accept), outputMessage);
      }
    }
  }
}

Answer 1

至少您应该在“ showToast”上添加一个“延迟”变量，并将Toast调用移到发布的Runnable内的中：

private fun showToast(text: String, delayInMilliseconds: Int) {
    val context = this
    handler.postDelayed (
        { Toast.makeText(context, text, Toast.LENGTH_SHORT).show() },
        delayInMilliseconds
    )
}

那你会喜欢它

showToast(cowName.toString(), 0)
showToast(cowWeight.toString(), 1000)
showToast(cowSex.toString(), 2000)
showToast(cowAge.toString(), 3000)

或执行该工作所需的任何延迟值。

试图一次显示烤面包消息

1 个答案: