Question

我正在使用以下keyUpDelay函数将keyup上的按键之间的代码执行延迟1秒。如何改变这一点，以便如果按下回车键，计时器将立即完成并执行代码。

var keyUpDelay = (function () { 

    var timer = 0; 

    return function (callback, ms) { 
        clearTimeout(timer); 
        timer = setTimeout(callback, ms); 
    }; 

})();


$('input').keyup(function () { // AUTO SEARCH FUNCTION

    keyUpDelay(function () {

        // code to delay by 1000 ms except if enter key is pressed

    }, 1000);

});

Answer 1

这个怎么样

keyUpDelay = (function () { 
 var timer = keyUpDelay.timer = 0; 
 return function (callback, ms) { 
  clearTimeout(timer); 
  timer = setTimeout(callback, ms); 
}; })(); 

$('input').keyup(function (e) { // AUTO SEARCH FUNCTION 
  if (e.which==13) {
   clearTimeout (keyUpDelay.timer);
   cb();
   return;
 }
 keyUpDelay(cb , 1000); 
 function cb () {
     // code to delay by 1000 ms except if enter key is pressed 
   }
});

除非按下回车键，否则延迟执行密钥上的代码

1 个答案: