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Css right top corner


Css right top corner

By : user2955437
Date : November 22 2020, 10:40 AM
around this issue Thank you for giving the rest of your HTML and CSS. The problem comes from the css line:
code :
.nav li {
    display: inline-block;
    margin-right: -5px;
}
.nav li:last-of-type {
    margin-right:0;    
}


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Given four Corner vertices of a rectangle, how to identify all upper and lower (left & right) corner points?

Given four Corner vertices of a rectangle, how to identify all upper and lower (left & right) corner points?


By : sdbern
Date : March 29 2020, 07:55 AM
this one helps. Four Corner Vertices of a deformed rectangle are given i.e. p1(x1,y1), p2(x2,y2), p3(x3,y3) & p4(x4,y4). , This is solution I worked out.
code :
fx=[x1 x2 x3 x4];  %represent x-cord
fy=[y1 y2 y3 y4];   %represent y-cord 
[xmn, ixmn]=min(fx);
fx(ixmn)=NaN;
[xmn2,ixmn2]=min(fx);
fx(ixmn)=xmn;   % to restore original data

if(fy(ixmn)>fy(ixmn2))
    ul=ixmn2; %upper-left
    ll=ixmn;  %lower left 
else
    ul=ixmn; 
    ll=ixmn2;
end
idd=setdiff(1:4,[ul ll]); 
if(fy(idd(1))>fy(idd(2)))
    ur=idd(2); %upper right
    lr=idd(1); %lower right
else
    ur=idd(1);
    lr=idd(2);
end
React Native maps get top left corner and bottom right corner latitude and longitude

React Native maps get top left corner and bottom right corner latitude and longitude


By : kamal debnath
Date : March 29 2020, 07:55 AM
wish help you to fix your issue If (REGION.latitude, REGION.longitude) is the center of the map and the deltas are the distance (in degrees) between the minimum and maximum lat/long displayed on the map then you should be able to get the topLeft and bottomRight points like below:
code :
            {
                latlng: {
                    latitude: REGION.latitude+(REGION.latitudeDelta/2),
                    longitude: REGION.longitude-(REGION.longitudeDelta/2)
                },
                title:'topLeft'

            },
            {
                latlng: {
                    latitude: REGION.latitude-(REGION.latitudeDelta/2),
                    longitude: REGION.longitude+(REGION.longitudeDelta/2)
                },
                title:'bottomRight'

            }
Not able to draw rectangle on Wpf canvas from lower right corner to upper left corner

Not able to draw rectangle on Wpf canvas from lower right corner to upper left corner


By : user7405658
Date : March 29 2020, 07:55 AM
this will help I am implementing functionality to allow user to draw rectangle on a Wpf canvas at run time by dragging mouse.I am currently able to able to draw the rectangle when I drag mouse from top left corner to bottom left, but the rectangle is not visible when I drag mouse from bottom left corner to top.Below is the xaml code that I am using: , Better use a Path with a RectangleGeometry:
code :
<Canvas Background="Transparent"
        MouseLeftButtonDown="CanvasContainer_MouseLeftButtonDown"
        MouseLeftButtonUp="CanvasContainer_MouseLeftButtonUp"
        MouseMove="CanvasContainer_MouseMove">
    <Path Stroke="Red">
        <Path.Data>
            <RectangleGeometry x:Name="selectionRect"/>
        </Path.Data>
    </Path>
</Canvas>
private Point? startPoint;

private void CanvasContainer_MouseLeftButtonDown(object sender, MouseButtonEventArgs e)
{
    var element = (UIElement)sender;
    element.CaptureMouse();
    startPoint = e.GetPosition(element);
}

private void CanvasContainer_MouseLeftButtonUp(object sender, MouseButtonEventArgs e)
{
    ((UIElement)sender).ReleaseMouseCapture();
    startPoint = null;
}

private void CanvasContainer_MouseMove(object sender, MouseEventArgs e)
{
    if (startPoint.HasValue)
    {
        selectionRect.Rect = new Rect(
            startPoint.Value, e.GetPosition((IInputElement)sender));
    }
}
Fabric Js: How to Create Rectangle programatically with only 2 points that is Top Left Corner and Bottom Right Corner?

Fabric Js: How to Create Rectangle programatically with only 2 points that is Top Left Corner and Bottom Right Corner?


By : an_xinwo
Date : March 29 2020, 07:55 AM
wish helps you The rectangle can be created using top, left, width, height like this , For your rectangle,
code :
width = difference between top-left x and right-bottom x,
height = difference between top-left y and right-bottom y.
var canvas = new fabric.Canvas('c');

function getValue(id) {
  return document.getElementById(id).value
}

function r() {
  return Math.floor(Math.random() * 255)
}

function getColor() {
  return 'rgb(' + r() + "," + r() + "," + r() + ')';
}

function addRect() {
  var tlx = getValue('tlx'),
    tly = getValue('tly'),
    brx = getValue('brx'),
    bry = getValue('bry');
  if (tlx == '' || tly == '' || brx == '' || bry == '') {
    alert('Enter all values');
    return false;
  }
  var width = Math.abs(+tlx - +brx),
    height = Math.abs(+tly - +bry);

  var rect = new fabric.Rect({
    left: +tlx,
    top: +tly,
    width: width,
    height: height,
    fill: getColor()
  });
  canvas.add(rect);
}
canvas{
 border: 1px solid #000;
}
<script src="https://rawgit.com/kangax/fabric.js/master/dist/fabric.js"></script>
Top-left x: <input type='number' min="0" max="300" id='tlx'><br>
Top-left y: <input type='number' min="0" max="300" id='tly'><br>
Bottom-Right x: <input type='number' min="0" max="300" id='brx'><br>
Bottom-Right x: <input type='number' min="0" max="300"id='bry'><br>
<button onclick='addRect()'>Add</button>
<canvas id="c" width="400" height="300"></canvas>
FAST corner detection implementation in MATLAB finding non-corner features

FAST corner detection implementation in MATLAB finding non-corner features


By : dOkI
Date : March 29 2020, 07:55 AM
it should still fix some issue This is one of the unfortunate difficulties with MATLAB syntax, and which often causes confusion.
As you know, MATLAB indexes into matrices as [row,column], which is natural in linear algebra and totally appropriate for matrices. But because MATLAB has expanded way beyond linear algebra, matrices are often used to store spatial data, in which x increases along the columns of the matrix, and y increases along the rows. Thus, a spatial location [x,y] should be reversed to find indices into the matrix: [y,x].
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