Question

我想创建简单的自动提交。例如，如果用户1秒没有在输入中键入任何内容或按下输入它提交。我的代码：

＆＃13;

$(document).on('keypress', $('input'), function(e) {
  console.log('clear')
  
  clearInterval(SubmitInterval);
  
  if (e.which == 13) {
    console.log('submit')
  }
  
  var SubmitInterval = setInterval(function() {
    console.log('submit')
    clearInterval(SubmitInterval);
  }, 1000);
});

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type=text id=1 />

＆＃13;

但它的工作很糟糕。感谢

Answer 1

问题是由于您在每个按键上重新定义SubmitInterval变量。而是在事件处理程序之外声明一次。试试这个：

＆＃13;

var submitInterval;

$(document).on('keypress', 'input', function(e) {
  console.log('clear')
  
  clearInterval(submitInterval);
  
  if (e.which == 13) {
    console.log('submit')
  }
  
  submitInterval = setInterval(function() {
    console.log('submit')
    clearInterval(submitInterval);
  }, 1000);
});

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type=text id=1 />

＆＃13;

Answer 2

现在一切正常。感谢。

var SubmitInterval;
$(document).on('keypress', $('input'), function(e) {
  clearInterval(SubmitInterval);
  SubmitInterval = setInterval(function () {
    console.log('submit');
    clearInterval(SubmitInterval);
  },1700);
  if(e.which == 13) {
	  console.log('submit');
	  clearInterval(SubmitInterval);
	};
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type=text />

通过JS自动提交表单（Timeout，TimeInterval）

2 个答案: