Border radius in SVG d3 - svg

I have following svg:
` svg.append('g')
.append("path")
.attr("d", "M 0 150 L 0 0 L 190 0 L 190 150 Z")
.attr("fill", "#f6f6f6"); `
Look like this:
But I want like this:
I don't have much knowledge about svg curve. How can I set border radius like this?

Done with Quadratic Bezier path:
svg {
background-color: #aaa;
}
<svg width="300" height="180">
<g transform="translate(10,10)">
<path d="M 60,0 H 180 Q 190,0 190,10 V 140 Q 190,150 180,150 H 10 Q 0,150 0,140 V 60 Q 0,0 60,0 Z" fill="#d6d6d6"/>
</g>
</svg>

You can find a detailed description of path here:
https://developer.mozilla.org/en-US/docs/Web/SVG/Tutorial/Paths
The simplest way to round a corner is Quadratic Bezier path.
For example, if you need to round a top-right corner with approximate radius r, and the coordinate of the corner are x and y, The path will look like this:
Move to (M) / Line to (L) / Horizontal line (H) to x - r, y
Quadratic Bezier (Q) to x, y + r via x, y
Line to (L) / Vertical line (V) to the next point
For example, if the radius is 20 and the corner is in the point 100,100 the path will be H 80,100 Q 100,100 100,120 V ...

Related

change 0,0 position of svg path

I have a svg path as below:
'M12,2 C 8.13,2 5,5.13 5,9 c 0,5.25 7,13 7,13s7,-7.75 7,-13C19,5.13 15.87,2 12,2zM7,9c0,-2.76 2.24,-5 5,-5s5,2.24 5,5c0,2.88 -2.88,7.19 -5,9.88C9.92,16.21 7,11.85 7,9z M12,9m-2.5,0a2.5,2.5 0,1 1,5 0a2.5,2.5 0,1 1,-5 0'
Now the tip point at (12,22), I would like to keep shape the same but the tip point at (0,0).
Which tool can do such position shift?
One possible solution would be using this tool to convert the path to to all-relative path commands.
For example convert:
M12,2 C 8.13,2 5,5.13 5,9 c 0,5.25 7,13 7,13s7,-7.75 7,-13C19,5.13 15.87,2 12,2zM7,9c0,-2.76 2.24,-5 5,-5s5,2.24 5,5c0,2.88 -2.88,7.19 -5,9.88C9.92,16.21 7,11.85 7,9z M12,9m-2.5,0a2.5,2.5 0,1 1,5 0a2.5,2.5 0,1 1,-5 0
to:
M12,2c-3.87,0,-7,3.13,-7,7c0,5.25,7,13,7,13s7,-7.75,7,-13c0,-3.87,-3.13,-7,-7,-7zm-5,7c0,-2.76,2.24,-5,5,-5s5,2.24,5,5c0,2.88,-2.88,7.19,-5,9.88c-2.08,-2.67,-5,-7.03,-5,-9.88zm5,0m-2.5,0a2.5,2.5,0,1,1,5,0a2.5,2.5,0,1,1,-5,0
Now in order to move it in the 0,0 just change the first two values after the initial M command from 12,2 to 0,-18.
The first value is 0 and represents the x value of both tue top and the tip. The second value is -18 and represents the y value of the starting point of the path located almost at the top. In order to get this value I did: 2 (actual y coordinate of the point) - 20 (height of the path).
In order to get the height of the path I'm using getBBox()
console.log(g.getBBox())
svg{width:30vh;border:solid}
<svg viewBox="-7 -18 14 20">
<path id="g" d="M0,-18c-3.87,0,-7,3.13,-7,7c0,5.25,7,13,7,13s7,-7.75,7,-13c0,-3.87,-3.13,-7,-7,-7zm-5,7c0,-2.76,2.24,-5,5,-5s5,2.24,5,5c0,2.88,-2.88,7.19,-5,9.88c-2.08,-2.67,-5,-7.03,-5,-9.88zm5,0m-2.5,0a2.5,2.5,0,1,1,5,0a2.5,2.5,0,1,1,-5,0"/>
</svg>

Is it possible to decompose a generic [a, b, c, d, e, f] (svg) transform matrix into a series of rotate/scale/translate directives?

This is not a duplicate. I've looked through all of the other answers.
I have an [a, b, c, d, e, f] svg transform matrix. I want to decompose it into any series of translate/scale/rotate(with optional center) operations. Skew is NOT an option. I'm trying to fit within the 7 attributes provided by the Android Vector Drawable Group (eg, rotation, pivotX, pivotY, scaleX, scaleY, translateX, translateY).
My first question is, is that possible for all such matrices? If the matrix has skew along either axis, can that be instead rendered via a series of rotate|scale operations? If not all matrices are possible, is it possible to detect when they're not?
The second question is some help with the basic math. I get as far as translateX = e and translateY = f. scaleX = a and scaleY = d IF b and c are zero. But when b and c are non zero, rotation and scale get entangled. How can I disentangle those?
Based on the answer here:
const sx = Math.sign(a) * Math.sqrt(a * a + c * c);
const sy = Math.sign(d) * Math.sqrt(b * b + d * d);
const tx = e;
const ty = f;
const angle = Math.atan2(b, d) * 180 / Math.PI;
const transform = `translate(${tx} ${ty}) scale(${sx} ${sy}) rotate(${angle})`;
It looks like there is no way to split the matrix into individual transforms when b / d != -a / c. Whether you can solve this using multiple rotations and scaling, I don't know.
<svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 400 250">
<path fill="blue" opacity="0.5" transform="translate(100 -150) scale(2.8284271247461903 0.9998489885977782) rotate(45)" d="M 238 162.4 247 214.7 200 190 153 214.7 162 162.4 123.9 125.3 176.5 117.6 200 70 223.5 117.6 276.1 125.3z"/>
<path fill="none" stroke="blue" transform="matrix(2 0.707 -2 0.707 100 -150)" d="M 238 162.4 247 214.7 200 190 153 214.7 162 162.4 123.9 125.3 176.5 117.6 200 70 223.5 117.6 276.1 125.3z" />
</svg>

Trouble creating a closed line chart in svg

I've got a react component which is creating an svg line chart (I'm not using a library, just creating the svg itself).
Problem is, when I add a final point to the path to return to the starting point, I have a strange 45degree angle showing up at the end of the chart.
Can anybody explain why this is not nicely closed?
Here's an example https://jsfiddle.net/7svavrmu/1/
From what I understand, the final L 0 300 should be returning the path to the origin.
Here's the code
<svg width="300" height="67.40652464075235">
<path fill="blue" stroke="black"
d="M 0 40.32613081539207
L 0.15306122448979592 40.990776224724726
L 0.25510204081632654 41.834373941621585
L 0.30612244897959184 62.31225269212592
L 299.0816326530612 45.84534164491692
L 299.33673469387753 65.256033885832
L 299.48979591836735 45.314084715607414
L 300 45.27080004137377 L 0 300 "></path>
</svg>
In SVG paths, each letter is an instruction and the following numbers are the coordinates for that instruction.
Your path ends at a strange location, L 0 300 is the bottom left location but way off the viewport, you need to "draw" the bottom part of your graph by removing the last instruction and adding L 300 67 (bottom right corner) and L 0 67 (bottom left corner). Putting it all together your path needs to look like this:
d="M 0 40.32613081539207
L 0.15306122448979592 40.990776224724726
L 0.25510204081632654 41.834373941621585
L 0.30612244897959184 62.31225269212592
L 299.0816326530612 45.84534164491692
L 299.33673469387753 65.256033885832
L 299.48979591836735 45.314084715607414
L 300 45.27080004137377 L 300 67 L 0 67"

How to calculate the y coordinate of a rectangle without transforms in Inkscape?

Imagine I have an Inkscape file with the following rectangle:
I want to calculate the Y coordinate reported by Inkscape (596.654).
How can I do it (manually) ?
I tried this:
The top of the page seems to have a Y coordinate of 744.
I subtract from that number the y coordinate of the rectangle in the XML editor (417) and its height (37) and get 744 - 417 - 37 = 290.
Note that the rectangle doesn't have any transforms and doesn't belong to a group.
Here's the simplified version of your SVG with only the relevant information included:
<svg width="297mm" height="210mm"
viewBox="0 0 1052.3622 744.09448">
<g transform="translate(0,-308.26772)">
<rect x="216.1537" y="417.34927"
width="385.25827" height="37.859257"
style="stroke-width:1;"
/>
</g>
</svg>
Despite what you thought, there is a transform in there (in the group).
SVG internal coordinates have their origin in the top left. Whereas Inkscape displays a converted value relative to the more normal origin at bottom left. The displayed value also takes into account the stroke width.
Your rectangle is drawn (internal coords) at
y = rectY + translateY
= 417.34927 - 308.26772
= 109.08155
The page has a height of 744.09448. So the displayed ccoordinate will be:
y = pageHeight - rectY - rectH - strokeWidth/2
= 744.09448 - 109.08155 - 37.859257 - 0.5
= 596.653673

Line chart animation in svg using jquery

I want to do line chart (path animation) individually not whole chart area. Please refer below JS fiddle
http://jsfiddle.net/2LEyb/
currently i have implemented the animation for whole chart area i.e. area rect initially zero and then i will increase the width step by step. it will affect other charts because column chart also with in chart area but it will do animation link line. Please refer below code snippet currently i have done.
doAnimation: function () {
var clipRect = $(this.chartObj.gSeriesEle).find("#" + this.chartObj.svgObject.id + "_ChartAreaClipRect");
$(clipRect).animate(
{ width: this.chartObj.model.m_AreaBounds.Width },
{
duration: 2000,
step: function (now, fx) {
$(clipRect).attr("width", now);
}
});
this.chartObj.model.Animation=false;
},
i need to get the individual "line" path rectangle and then i want to increase the width of individual line path rectangle step by step in jquery animate instead increasing the width of entire chart area.
how to calculate the rectangle for individual path and then increase the width of rectangle.
<g id="container_svg_SeriesGroup_0" transform="translate(144,432)"><path id="container_svg_John_0" fill="none" stroke-width="3" stroke="url(#container_svg_John0Gradient)" stroke-linecap="butt" stroke-linejoin="round" d="M -2454.7999999999997 -125.8888888888889 L 0 0 M 0 0 L 258.40000000000003 -45.77777777777778 M 258.40000000000003 -45.77777777777778 L 516.8000000000001 -11.444444444444445 M 516.8000000000001 -11.444444444444445 L 646 -183.11111111111111 "/></g>
Thanks,
Siva

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