通過RaphaelJS套件循環

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我試圖通過一系列RaphaelJS套件循環。通過RaphaelJS套件循環

我有一系列的箭頭形狀（套）。我希望能夠通過它們循環。

var migration1 = c.set(c.path("M129.045,368.584c0,0,19.155-13.9,23.374-50.593 c2.283-19.858-10.22-31.366-18.958-36.886").attr({fill: "none", "stroke-width": 3, stroke: "#990000"}), c.path("M139.728,278.292L135.038 282.091 134.516 288.104 125.493 277.021 z").attr({fill: "#990000", "stroke-width": 0}), c.path("M136.059,375.577c0,0,42.113-48.3,94.441-84.818 c55.985-39.072,143.717-72.937,171.049-83.015").attr({fill: "none", "stroke-width": 3, stroke: "#990000"}), c.path("M399.536,214.469L399.827 208.441 395.689 204.047 409.966 204.699 z").attr({fill: "#990000", "stroke-width": 0})); 
var migration2 = c.set(c.path("M287.563,407.934 c0,0,26.391-73.864,66.459-135.858c30.304-46.886,62.735-75.691,76.743-86.947").attr({fill: "none", "stroke-width": 3, stroke: "#990000"}), c.path("M430.854,192.14L429.326 186.301 424.063 183.347 437.881 179.696 z").attr({fill: "#990000", "stroke-width": 0})); 
var migration3 = c.set(c.path("M106.872,237.5c0,0,82.633-29.237,261.38-61.119 c145.376-25.929,228.743-26.742,254.803-26.142").attr({fill: "none", "stroke-width": 3, stroke: "#990000"}), c.path("M618.66,155.717L621.198 150.242 619.018 144.614 632 150.589 z").attr({fill: "#990000", "stroke-width": 0}), c.path("M106.872,237.5c0,0,95.407-26.3,178.93-25.373 c62.451,0.693,87.378,29.095,95.981,43.382").attr({fill: "none", "stroke-width": 3, stroke: "#990000"}), c.path("M374.787,254.352L380.807 253.922 384.677 249.291 385.73 263.543 z").attr({fill: "#990000", "stroke-width": 0})); 
var migration4 = c.set(c.path("M336,292 c11.139-4.582,18.674-13.417,31.977-14.868c8.351-0.911,34.987-3.037,34.05,10.528c-0.977,14.141-32.223,18.288-32.963,30.188 c-0.384,6.178,9.04,9.792,12.936,7.151 M387,351 c0.673-0.509,1.006-1.176,1-2").attr({fill: "none", "stroke-width": 3, stroke: "#990000"})); 
var migration5 = c.set(c.path("M117.073,329.436 c0,0,12.897-27.418,68.927-67.291c38.271-27.234,71.998-42.664,102.902-52.111").attr({fill: "none", "stroke-width": 3, stroke: "#990000"}), c.path("M286.465,216.615L287.142 210.618 283.295 205.968 297.5 207.535 z").attr({fill: "#990000", "stroke-width": 0}));

我試過的變化：

for (i=1<=5;i++){ 
    this['migration'+i].hide(); 
}

......這樣我就可以有一個按鈕，顯示它們依次是：

var current=1; 
$('#next').click(function(){ 
    this['migration'+current].hide(); 
    current++; 
    this['migration'+current].show(); 
})

使用 '這個' 沒有按看起來不錯，但我不知道如何連接設置名稱來定位它。

非常感謝提前。

來源

2013-02-28 Brian Williamson

您想嘗試將所有集合放入更大的集合中，然後使用forEach（）方法嗎？我在想這就是你要找的東西。 – tnsingle 2013-02-28 17:53:03

您可以創建套陣列...

var migrations =[ c.set(...), c.set(...), c.set(...), c.set(...), c.set(...)];

那麼你的代碼看起來像：

var current=1; 
$('#next').click(function(){ 
    migrations[current].hide(); 
    migrations[current++].show(); 
})

來源

2013-03-01 15:25:13 Duopixel

我用下面的方法，例如隱藏了一組傳遞到功能爲 '文本'：

function hideSet(set) { 
    set=eval(set); 
    set.forEach(function(e){ e.hide(); }) 
}

使用此來切換

hideSet('setname');

來源

2013-10-15 09:55:23 Risingson

通過RaphaelJS套件循環

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