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How do you track each setInterval cycle?

开发者 https://www.devze.com 2023-02-16 18:36 出处:网络
function count(){ //user input stored in the last index of the array each cycle array[lastindex]=userinput;
function count(){
    //user input stored in the last index of the array each cycle
    array[lastindex]=userinput;
    setInterval(count,5000);
}

Like me sample codes above, I am populating the array by storing the user inputs.However, I want to add one more case to this, which is if the users dont type in anything within 5 seconds, I am still able to store a "mark" that indicates that. in other words, how can I detect this automatically before t开发者_StackOverflow中文版he next cycle? Thank you


I imagine it like this - store a variable that indicates that user is typing now. Add an event, for example, keyup to the input where user types in, that after 5 sec of idleing changes that varianble.

var
   userTyping = false,
   typingId;


$("input").keyup(function(){
   userTyping = true;
   window.clearInterval(typingId);
   typingId = window.setTimeout(function(){userTyping = false;}, 5000);
});

And then the check:

function count(){
//user input stored in the last index of the array each cycle
if(!userTyping){
  array[lastindex]=userinput;
}
setTimeout(count,5000);
}

Note that you should use setTimeout if within the function, not the setInterval each time you call - that would end up badly...


So first question is does no new input means userInputOld === userInput or userInput === ""? I'll guess no, because than it would be trivial to check.

You must make a variable that will mark whether there was input or not. Let's presume keyboard input:

var hasInput = false;
$("input").keyup(function(){hasInput = true;})

Then you can check it in your code

function count(){
    //user input stored in the last index of the array each cycle
    if (hasInput) {
        array[lastindex]=userinput;
    } else {
        aray[lastindex]=null;
    }
    setInterval(count,5000);
}

Anyway I would like to ask what exactly are you doing, because it looks like you can probably do it better altogether.

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