te')); return $arr; } /* 遍历用户所有主题 * @param $uid 用户ID * @param int $page 页数 * @param int $pagesize 每页记录条数 * @param bool $desc 排序方式 TRUE降序 FALSE升序 * @param string $key 返回的数组用那一列的值作为 key * @param array $col 查询哪些列 */ function thread_tid_find_by_uid($uid, $page = 1, $pagesize = 1000, $desc = TRUE, $key = 'tid', $col = array()) { if (empty($uid)) return array(); $orderby = TRUE == $desc ? -1 : 1; $arr = thread_tid__find($cond = array('uid' => $uid), array('tid' => $orderby), $page, $pagesize, $key, $col); return $arr; } // 遍历栏目下tid 支持数组 $fid = array(1,2,3) function thread_tid_find_by_fid($fid, $page = 1, $pagesize = 1000, $desc = TRUE) { if (empty($fid)) return array(); $orderby = TRUE == $desc ? -1 : 1; $arr = thread_tid__find($cond = array('fid' => $fid), array('tid' => $orderby), $page, $pagesize, 'tid', array('tid', 'verify_date')); return $arr; } function thread_tid_delete($tid) { if (empty($tid)) return FALSE; $r = thread_tid__delete(array('tid' => $tid)); return $r; } function thread_tid_count() { $n = thread_tid__count(); return $n; } // 统计用户主题数 大数量下严谨使用非主键统计 function thread_uid_count($uid) { $n = thread_tid__count(array('uid' => $uid)); return $n; } // 统计栏目主题数 大数量下严谨使用非主键统计 function thread_fid_count($fid) { $n = thread_tid__count(array('fid' => $fid)); return $n; } ?>javascript - How to get info on what key was pressed on for how long? - Stack Overflow
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javascript - How to get info on what key was pressed on for how long? - Stack Overflow

programmeradmin3浏览0评论

Imagine this code:

if (navigator.appName == "Opera")
    document.onkeypress = function (e) { console.log(e.keyCode); };
else 
    document.onkeydown = function (e) { console.log(e.keyCode); };

What it does is pretty obvious, I guess. The problem is, if you hold the key for a long period of time, it will be registered a lot of times. In my app, this is a problem because it makes my app do a lot of unnecessary puting. Is it possible to somehow only get a keypress once, but with info on how long the key was held?

Imagine this code:

if (navigator.appName == "Opera")
    document.onkeypress = function (e) { console.log(e.keyCode); };
else 
    document.onkeydown = function (e) { console.log(e.keyCode); };

What it does is pretty obvious, I guess. The problem is, if you hold the key for a long period of time, it will be registered a lot of times. In my app, this is a problem because it makes my app do a lot of unnecessary puting. Is it possible to somehow only get a keypress once, but with info on how long the key was held?

Share Improve this question edited Jun 1, 2020 at 16:53 Brian Tompsett - 汤莱恩 5,89372 gold badges61 silver badges133 bronze badges asked Jun 24, 2011 at 19:17 Aleksejs PopovsAleksejs Popovs 8702 gold badges12 silver badges18 bronze badges 3
  • I'm no expert on logic or anything but shouldn't the branches of an if statement differ? :) – Šime Vidas Commented Jun 24, 2011 at 19:59
  • Do you use a library like jQuery? – Šime Vidas Commented Jun 24, 2011 at 20:03
  • Oops, yes, you're right, of course they should differ %) No, I don't use jQuery or any other framework. – Aleksejs Popovs Commented Jun 25, 2011 at 7:47
Add a ment  | 

2 Answers 2

Reset to default 13

Here you go:

var pressed = {};

window.onkeydown = function(e) {
    if ( pressed[e.which] ) return;
    pressed[e.which] = e.timeStamp;
};

window.onkeyup = function(e) {
    if ( !pressed[e.which] ) return;
    var duration = ( e.timeStamp - pressed[e.which] ) / 1000;
    // Key "e.which" was pressed for "duration" seconds
    pressed[e.which] = 0;
};

Live demo: http://jsfiddle/EeXVX/1/show/

(remove the "show/" part of the URL to view the code for the demo)

So you have the pressed object which monitors which keys are currently being pressed and at what point (in time) they have been pressed.

Inside the keyup handler, you determine if the key was being pressed, and if so, calculate the duration by subtracting the time-stamps of the keyup/keydown events.

Have you tried doing somethign like,

  1. onkeydown, remove keydown listener.
  2. onkeyup, attach keydown listener again and puter the time?
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