ui-router中的錨點問題（angular中的錨點問題）

angular.module('anchorScrollExample', []) .controller('ScrollController', ['$scope', '$location', '$anchorScroll', function($scope, $location, $anchorScroll) { $scope.gotoBottom = function() { // set the location.hash to the id of // the element you wish to scroll to. $location.hash('bottom'); // call $anchorScroll() $anchorScroll(); }; }]); 官網文檔傳送門：https://docs.angularjs.org/api/ng/service/$anchorScroll 綜上，angular對錨點作了處理，使用時先用$location.hash(),指定位置，hash的參數就是錨點id，而後再使用$anchorScroll()移動到錨點；


若是對用戶體驗有必定要求，則可能須要在跳轉的過程當中出現滾動的效果，angular中並無提供這樣的api，不過能夠本身對照anchorScroll自定義，下面就是一個例子，能夠直接使用到項目中

app.service('anchorSmoothScroll', function(){

this.scrollTo = function(eID) {

// This scrolling function
// is from http://www.itnewb.com/tutorial/Creating-the-Smooth-Scroll-Effect-with-JavaScript

var startY = currentYPosition();
var stopY = elmYPosition(eID);
var distance = stopY > startY ? stopY - startY : startY - stopY;
if (distance < 100) {
scrollTo(0, stopY); return;
}
var speed = Math.round(distance / 100);
if (speed >= 20) speed = 20;
var step = Math.round(distance / 25);
var leapY = stopY > startY ? startY + step : startY - step;
var timer = 0;
if (stopY > startY) {
for ( var i=startY; i<stopY; i+=step ) {
setTimeout("window.scrollTo(0, "+leapY+")", timer * speed);
leapY += step; if (leapY > stopY) leapY = stopY; timer++;
} return;
}
for ( var i=startY; i>stopY; i-=step ) {
setTimeout("window.scrollTo(0, "+leapY+")", timer * speed);
leapY -= step; if (leapY < stopY) leapY = stopY; timer++;
}

function currentYPosition() {
// Firefox, Chrome, Opera, Safari
if (self.pageYOffset) return self.pageYOffset;
// Internet Explorer 6 - standards mode
if (document.documentElement && document.documentElement.scrollTop)
return document.documentElement.scrollTop;
// Internet Explorer 6, 7 and 8
if (document.body.scrollTop) return document.body.scrollTop;
return 0;
}

function elmYPosition(eID) {
var elm = document.getElementById(eID);
var y = elm.offsetTop;
var node = elm;
while (node.offsetParent && node.offsetParent != document.body) {
node = node.offsetParent;
y += node.offsetTop;
} return y;
}

};

});

原址：http://jsfiddle.net/brettdewoody/y65G5/