I have the following code to get a feature from a .geojson, and color the polygons (which are squares in my case) according to values in the .geojson file.
var myStyle = function(feature) {
var id = feature.get('featureName');
fill.setColor(id >= 0 && id <= 0.10 ? coloringFunction1(feature): id >= 0.10 && id <= 1.00 ? coloringFunction2(feature):coloringFunction3(feature));
return style;
};
I want to show the id value as well, along with the colors, so I have done like the following:
var myStyle = function(feature) {
var id = feature.get('featureName');
fill.setColor(id >= 0 && id <= 0.10 ? coloringFunction1(feature): id >= 0.10 && id <= 1.00 ? coloringFunction2(feature):coloringFunction3(feature));
// To show the text
text: new ol.style.Text({
textAlign: "Center",
textBaseline: "Middle",
font: 'Normal 12px Arial',
text: id,
fill: new ol.style.Fill({
color: 'black'
}),
stroke: new ol.style.Stroke({
color: 'black',
width: 3
}),
})
return style;
};
However, it's not working at all. So my question is: How to show the extracted feature value with its color?
Related
My gpx file already contains elevation information for each trkpt and I can augment this with a speed for each trkpt. I would like to represent the elevation or the speed at each trkpt by varying the color of the track. For instance: slow is blue, fast is red.
How can I do this?
And this probably means: Which files and functions in Openlayers do I have to change to do this?
You can try the ol/style/FlowLine of ol-ext to achieve this.
Using this style, you can change the with/color of the feature along the line using a function. This example show how to: http://viglino.github.io/ol-ext/examples/style/map.style.flowline2.html.
You just have to calculate the width (or color) along the feature geometry varying according the speed or altitude:
const flowStyle = new ol.style.FlowLine({
width: function(f, step) {
// calculate the with of the feature f at the given step
// step is the curvilinear abscissa between 0,1
// (0: first coordinate, 1: last one)
const width = ...
return width;
}
});
#+
You should go with a stylefunction for the vector layer:
https://openlayers.org/en/v4.6.5/apidoc/ol.html#.StyleFunction
This function is checked for each feature to be displayed on the vector layer and the related style can be set/returned programmatically. For example:
function gpxStyle(feature) {
var style = null;
if (feature.get("speed")>="100") {
style = new ol.style.Style({
image: new ol.style.Circle({
radius: 6,
stroke: new ol.style.Stroke({
color: 'red',
width: 2
}),
fill: new ol.style.Fill({
color: 'red'
})
})
});
}
else {
style = new ol.style.Style({
image: new ol.style.Circle({
radius: 6,
stroke: new ol.style.Stroke({
color: 'blue',
width: 2
}),
fill: new ol.style.Fill({
color: 'blue'
})
})
});
}
return [style];
}
var gpxLayer = new ol.layer.Vector({
source: new ol.source.Vector(),
style: gpxStyle
});
How can I get canvas-relative position (top, left) of triangle inside an group as bellow image?
I followed this topic: How to get the canvas-relative position of an object that is in a group? but it only right when group is not rotated.
Working example you may find here: http://jsfiddle.net/mmalex/2rsevdLa/
Fabricjs provides a comprehensive explanation of how transformations are applied to the objects: http://fabricjs.com/using-transformations
Quick answer: the coordinates of an object inside the group is a point [0,0] transformed exactly how the object in the group was transformed.
Follow my comments in code to get the idea.
// 1. arrange canvas layout with group of two rectangles
var canvas = new fabric.Canvas(document.getElementById('c'));
var rect = new fabric.Rect({
width: 100,
height: 100,
left: 50,
top: 50,
fill: 'rgba(255,0,0,0.25)'
});
var smallRect = new fabric.Rect({
width: 12,
height: 12,
left: 150 - 12,
top: 50 + 50 - 12 / 2 - 10,
fill: 'rgba(250,250,0,0.5)'
});
// 2. add a position marker (red dot) for visibility and debug reasons
var refRect = new fabric.Rect({
width: 3,
height: 3,
left: 100,
top: 100,
fill: 'rgba(255,0,0,0.75)'
});
var group = new fabric.Group([rect, smallRect], {
originX: 'center',
originY: 'center'
});
canvas.add(group);
canvas.add(refRect);
canvas.renderAll();
// 3. calculate coordinates of child object in canvas space coords
function getCoords() {
// get transformation matrixes for object and group individually
var mGroup = group.calcTransformMatrix(true);
// flag true means that we need local transformation for the object,
// i.e. how object is positioned INSIDE the group
var mObject = smallRect.calcTransformMatrix(true);
console.log("group: ", fabric.util.qrDecompose(mGroup));
console.log("rect: ", fabric.util.qrDecompose(mObject));
// get total transformattions that were applied to the child object,
// the child is transformed in following order:
// canvas zoom and pan => group transformation => nested object => nested object => etc...
// for simplicity, ignore canvas zoom and pan
var mTotal = fabric.util.multiplyTransformMatrices(mGroup, mObject);
console.log("total: ", fabric.util.qrDecompose(mTotal));
// just apply transforms to origin to get what we want
var c = new fabric.Point(0, 0);
var p = fabric.util.transformPoint(c, mTotal);
console.log("coords: ", p);
document.getElementById("output").innerHTML = "Coords: " + JSON.stringify(p);
// do some chores, place red point
refRect.left = p.x - 3 / 2;
refRect.top = p.y - 3 / 2;
canvas.bringToFront(refRect);
canvas.renderAll();
}
a very simple way to get topleft is
var cords = object._getLeftTopCoords();
cords.x and cord.y will give you the result
I have a problem in Rappid/jointJS
I have in stencil.js 4 shapes(2 basic.Circle and 2 basic.Rect) with names START(basic.Circle), END(basic.Circle), Activity(basic.Rect) and Workitem( basic.Rect) and I want in my main.js from all my graph to get the basic shape with name(I mean with attrs text ) "Activity".
This is the Stencil description for "Activity" :
new joint.shapes.basic.Rect({ size: { width: 5, height: 3 },
attrs: {
rect: {
rx: 2, ry: 2, width: 50, height: 30,
fill: '#0000FF'
},
text: { text: 'Activity', fill: '#ffffff', 'font-size': 10,
stroke: '#000000', 'stroke-width': 0 }
}
}),
How wil I get it? The only way I can search in my graph so far is if a cell has type basic.Circle(use of get('type') === 'basic.Circle')). but with type Circle I have two items:Activity and Workitem.
Is it so difficult to search for the graph element with name : "Activity"?
Thank you in advance
You can obtain all the elements (except for links) from following method
var allElement = graph.getElements()
Next if you want to obtain elements with 'Activity' do as follows
var activityElements = [];
allElement.forEach(elem => {
var textVal = elem.attributes.attrs.text.text;
if(textVal !== undefined && textVal === 'Activity') {
activityElements.push(elem);
}
});
Now the activityElements array will contain all the elements you require.
I solved my problem by taking element data in JSON format:
_.each(this.graph.getElements(), function(element) {
if(element.attributes.attrs["text"]["text"] == "Activity"){
//alert("YEAHHHHHH");
}
});
you could use the api on element as well, element.attr('text') returns the text object from the shape: { text: 'Activity', fill: '#ffffff', 'font-size': 10,
stroke: '#000000', 'stroke-width': 0 }
You could also set an "id" attribute to your shape and use graph.getCell('id_name_goes_here'); which would be much simpler if you didn't mind adding an id field to each shape.
I'm using Appcelerator Titanium and want to make an Android application. I created a TableView with 4 rows. I want to put 3 labels into each row and I want my labels to be evenly distributed. (1st label must be at the left, 2nd at the center, and 3rd at the right of the row.)
Thank you.
For left, center, right, you can use relative positioning and text alignment to very simply make your rows. This approach works well regardless of how wide the current screen is (ie, this works on tablet, phone, TV, etc).
var win = Ti.UI.createWindow({
backgroundColor: '#fff'
});
var rows = [];
for (var i = 0; i < 10; i++) {
var row = Ti.UI.createTableViewRow();
row.add(Ti.UI.createLabel({
text: 'Left ' + i, textAlign: 'left',
color: '#000',
left: 10
}));
row.add(Ti.UI.createLabel({
text: 'Center ' + i, textAlign: 'center',
color: '#000'
}));
row.add(Ti.UI.createLabel({
text: 'Right ' + i, textAlign: 'right',
color: '#000',
right: 10
}));
rows.push(row);
}
win.add(Ti.UI.createTableView({
data: rows
}));
win.open();
Another option would be to use percent widths, like left: '0%', width: '33%', then left: '33%', width: '33%', etc.
Or you could say the first label is from left: 0, width: 200. The second is left: 200, width: 50, and the third is from left: 250, right: 0. That would give you a third label that is elastic so it can take up al the space.
Yet another option (that I don't recommend you take) would be to use Ti.Platform.displayCaps.platformWidth and position the row elements based on that. But that would be very fragile to orientation changes.
It all depends on your content. With these in hand, you should be able to handle your particular use case.
Here is how I recently solved this issue. With 3 separate views and labels inside each row. This worked very well for me! All of them are based on percents, so it should work on all resolutions. Good luck!
for (var i = 0; i < 4; i++) {
var row = Ti.UI.createTableViewRow({
height: 'auto',
});
var view1 = Ti.UI.createView({
left : 0,
width : "33.33%",
backgroundColor:'red',
height:40
});
var label1 = Ti.UI.createLabel({
text: 'here',
color:"#fff",
textAlign: Ti.UI.TEXT_ALIGNMENT_LEFT
});
view1.add(label1);
var view2 = Ti.UI.createView({
left : "33.33%",
width : "33.33%",
backgroundColor : "white",
height:40
});
var label2 = Ti.UI.createLabel({
text: 'there',
color:"#fff"
});
view2.add(label2);
var view3 = Ti.UI.createView({
left : "66.66%",
width : "33.33%",
backgroundColor: "blue",
height:40
});
var label3 = Ti.UI.createLabel({
text: 'Everywhere',
color:"#fff"
});
view3.add(label3);
row.add(view1);
row.add(view2);
row.add(view3);
}
try this code.
var win = Ti.UI.createWindow({
});
var table = Ti.UI.createTableView({
backgroundColor:'blue',
top:50,
height : 160
});
win.add(table);
var data = [];
for (var i = 0; i < 4; i++) {
var row = Ti.UI.createTableViewRow({
height : 40
});
var label1 = Ti.UI.createLabel({
text : 'label1',
top : 5,
left : 10,
width : 80,
height : 30
});
row.add(label1);
var label2 = Ti.UI.createLabel({
text : 'label2',
top : 5,
left : 130,
width : 80,
height : 30
});
row.add(label2);
var label3 = Ti.UI.createLabel({
text : 'label3',
top : 0,
left : 240,
width : 80,
height : 30
});
row.add(label3);
data.push(row);
}
table.data = data;
win.open();
Best luck..
I'm trying to determine the length of a String so I can dynamically layout some Labels. In iOS I would use:
CGSize size = [string sizeWithFont:font];
Is it possible to do this with the Appcelerator API or is there another way to do this type of dynamic layout?
If you create labels with 'auto' width, once they've been constructed you can access their actual width and position any other labels accordingly:
var window = Ti.UI.createWindow();
var label = Ti.UI.createLabel({
text: 'This is a test string',
width: 'auto',
color: '#FFF',
left: 0,
top: 0
});
var labelWidth = label.width;
Ti.API.info(labelWidth);
// Dynamically position the second label
var secondLabel = Ti.UI.createLabel({
text: 'Second label',
width: 100,
left: labelWidth + 10,
top: 0,
color: '#FFF'
});
window.add(label);
window.add(secondLabel);
window.open();
It's not very elegant but it works.
var win = Ti.UI.createWindow({
'layout': 'horizontal'
});
https://github.com/appcelerator/titanium_mobile/blob/master/demos/KitchenSink/Resources/examples/horizontal_layout.js