以下测试基本上是〜1000次数学运算,并且在大多数PC和Android浏览器以及iOS 4.x上运行良好.在iOS5 safari(iPhone 4和iPad 2)上,我们得到"JavaScript:错误未定义JavaScript执行超出超时".任何帮助非常感谢谢谢.
/** Converts numeric degrees to radians */
if (typeof (Number.prototype.toRad) === "undefined") {
Number.prototype.toRad = function () {
return this * Math.PI / 180;
}
}
function gc(lat1, lon1, lat2, lon2) {
// returns the distance in km between a pair of latitude and longitudes
var R = 6371; // km
var dLat = (lat2 - lat1).toRad();
var dLon = (lon2 - lon1).toRad();
var a = Math.sin(dLat / 2) * Math.sin(dLat / 2) +
Math.cos(lat1.toRad()) * Math.cos(lat2.toRad()) *
Math.sin(dLon / 2) * Math.sin(dLon / 2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
var d = R * c;
return d;
}
function test() {
var d1 = new Date();
var lat1, lon1, lat2, lon2;
lat1 = -36;
lon1 = 174;
lat2 = lat1;
lon2 = lon1;
while (lat2 > -37) {
lat2 = lat2 - 0.001;
var stest = "lat1=" + lat1 + ",lon1=" + lon1 + ",lat2=" + lat2 + ",lon2=" + lon2 + "=" + gc(lat1, lon1, lat2, lon2);
}
var d2 = new Date();
var stest = (d2.getTime() - d1.getTime()) / 1000.0 + "s";
$("#lblTest").html(stest + "<BR/>" + $("#lblTest").html());
}
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jfr*_*d00 10
如果你想在javascript中执行一个长时间运行的操作,而你正在接近某些浏览器强制执行的脚本执行时间限制,那么你将不得不将你的功能分成多个部分,运行一个部分,到很短的时间setTimeout(fn, 1)然后执行下一个部分等...以这种方式执行,您可以运行代码数小时,因为它为其他脚本和其他事件提供了处理的机会.它有时需要少量的代码重构才能做到这一点,但总是可以做一点工作.
伪代码的基本概念是这样的:
var state = {}; // set initial state
var done = false;
function doWork() {
// do one increment of work that will never get even close to the browser
// execution time limit
// update the state object with our current operating state for the next execution
// set done = true when we're done processing
if (!done) {
setTimeout(doWork, 1);
}
}
doWork();
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在您的特定代码中,您可以执行类似的操作.您可以一次处理100个纬度点,然后执行短的setTimeout以执行下一个100,依此类推.您可以将该100个数字调整为最适合的数字.数字越大,您在每个计时器上执行的操作越多,总体执行时间越长,但您越接近浏览器脚本执行限制.setTimeout使浏览器保持活动状态(处理其他事件)并阻止执行时间限制.
function test() {
var d1 = new Date();
var lat1, lon1, lat2, lon2, done = false;;
lat1 = -36;
lon1 = 174;
lat2 = lat1;
lon2 = lon1;
function calcGC() {
var cntr = 0;
while (lat2 > -37 && cntr < 100) {
lat2 = lat2 - 0.001;
var stest = "lat1=" + lat1 + ",lon1=" + lon1 + ",lat2=" + lat2 + ",lon2=" + lon2 + "=" + gc(lat1, lon1, lat2, lon2);
cntr++;
}
// if we have more to go, then call it again on a timeout
if (lat2 > -37) {
setTimeout(calcGC, 1);
} else {
var d2 = new Date();
var stest = (d2.getTime() - d1.getTime()) / 1000.0 + "s";
$("#lblTest").html(stest + "<BR/>" + $("#lblTest").html());
}
}
calcGC();
}
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