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我正在构建一个应用程序，用户可以在其中无限次操作 svg 对象（例如图像），即旋转和缩放它们。我使用Raphael.js来处理 SVG。

如何在应用新转换之前将对象“重置”为其初始状态，以便新转换不会与前一个转换重叠？

Bellow 是一个测试（这是一个工作小提琴：jsfiddle.net/5TsW6/3/），我尝试在第二个视口中重置对象，然后再应用与第一个相同的转换，但我的尝试没有效果，因为两个视口中的结果是不同的，当它应该是相同的时：

<div id="wrapper1" style="width:250px; height:250px; outline:1px solid invert; float:left; margin-right:5px;"></div>
<div id="wrapper2" style="width:250px; height:250px; outline:1px solid invert; float:left;"></div>​

和脚本：

var img_width=100, img_height=75,
    canvas1, canvas2,
    image1, image2,
    w1, w2, // contours
    x,y,    // attributes "x" and "y"
    rx,ry;  // coordinates for the rotation pivot

// Create the canvases
canvas1 = new Raphael(document.getElementById("wrapper1"), 250, 250);  
canvas2 = new Raphael(document.getElementById("wrapper2"), 250, 250);


// Add the images to the canvases    
image1 = canvas1.image("http://picselbocs.com/projects/cakemyface/image-gallery/intermediate/3295094303_2b50e5fe03_z.jpg",0,0,img_width,img_height);
image2 = canvas2.image("http://picselbocs.com/projects/cakemyface/image-gallery/intermediate/3295094303_2b50e5fe03_z.jpg",0,0,img_width,img_height);


// Modify (x,y) 
x = 40;
y = 80;
image1.attr({"x":x,"y":y});
image2.attr({"x":x,"y":y});


// Add the objects' contours
canvas1.rect(x,y,img_width,img_height).attr({"stroke-width":"1px","stroke":"#00F"});
canvas2.rect(x,y,img_width,img_height).attr({"stroke-width":"1px","stroke":"#00F"});

// Compute the rotation pivot coordinates (the pivot should coincide with the object's center)
var scaling = 1.2;
rx = (x + img_width / 2) * scaling;
ry = (y + img_height / 2) * scaling;

// Apply transformations
image1.transform("S"+scaling+","+scaling+",0,0R45,"+rx+","+ry);
image2.transform("S"+scaling+","+scaling+",0,0R45,"+rx+","+ry);



// ----- Here starts second transformation ------ //
image2.transform("").transform("S"+scaling+","+scaling+",0,0r45,"+rx+","+ry);   // I've attempted to reset the transform string, but with no effect
​
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1 回答 1

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重置正在工作。您在第二次转换中打错了，使用r45而不是R45，意思是“在应用之前考虑之前的转换”，因此有所不同。

于 2012-10-28T10:26:50.943 回答