[英]loadJSON with lat and lon and get the variables
I am loading JSON file and I want to take the latitude and longitude variables from file.我正在加载 JSON 文件,我想从文件中获取纬度和经度变量。 Am doing it with let lat = row.get('lat');
我用let lat = row.get('lat'); but it doesn't work and in the const pix = yMap.latLngToPixel(data.lat,data.lon);
但它不起作用并且在 const pix = yMap.latLngToPixel(data.lat,data.lon); it says that it doesn't recognize it .
它说它不认识它。 How can I do it to work right, 'cause after that I want to take the json file from localhost, this is why I am using a p5.js library.
我如何才能正常工作,因为之后我想从本地主机获取 json 文件,这就是我使用 p5.js 库的原因。
<title>Mappa </title>
<script src="https://cdnjs.cloudflare.com/ajax/libs/p5.js/0.5.16/p5.min.js" type="text/javascript"></script>
<script src="https://cdn.jsdelivr.net/npm/mappa-mundi@0.0.4" type="text/javascript"></script>
<script src="mappa.js" ></script>
Json.file (like the one that I will take from localhost) Json.file(就像我从 localhost 获取的那个)
{"conflicts":[
{ "Place":"Afganistan",
"lat":33.0000,
"lon":65.0000 }
,
{"Place": "Koln",
"lat":17.0500,
"lon":-61.8000}
,
{"Place": "ai",
"lat":18.2500,
"lon": -63.166}
]
}
}
js.file js.文件
let data;
let myMap;
let canvas;
const mappa = new Mappa ('Leaflet');
const options = {
lat: 39.074207,
lng: 21.824312,
zoom: 2,
style: "http://{s}.tile.osm.org/{z}/{x}/{y}.png"
}
function preload(){
data = loadJSON ("conflicts.json");
//var jsonData = JSON.parse(data);
}
function setup(){
canvas = createCanvas(1350,600);
myMap = mappa.tileMap(options);
myMap.overlay(canvas);
let lat = row.get('lat');
let lon = row.get('lon');
}
function draw() {
clear();
const pix = myMap.latLngToPixel(data.lat,data.lon);
fill (255,0,200);
ellipse (pix.x,pix.y,20,20);
}
Your data
is a an object with a single property which is an array.您的
data
是一个具有单个属性的对象,该属性是一个数组。 To access each element in this array, you need to iterate it:要访问此数组中的每个元素,您需要对其进行迭代:
for (let item of data.conflicts)
draw(item)
function draw(item) {
const pix = myMap.latLngToPixel(item.lat, item.lon);
fill (255,0,200);
ellipse (pix.x,pix.y,20,20);
}
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