Question

我想计算一个圆圈的坐标。我不能让它打印坐标，我不确定它是否计算它们。我的代码：

HTML：

<p>Here is the coordinates: </p>

JS：

#screen size will use screen.width and screen.height later
var sW = 1920;
var sH = 1080;

#Two arrays with the coordinates stored in them
var XcircleCoordinates;
var YcircleCoordinates;

#rows should be rows in database. I will use php to get the information, but assume 4 for now.
var rows = 4;

#radius and center of circle.
var radius = 200;
var center = [sW/2,sH/2];

function xCord(i){
  XcircleCoordinates[i] = radius*Math.cos((2.0*Math.PI*i)/rows)+center[0];
  return XcircleCoordinates[i];
}

function yCord(i){
  YcircleCoordinates[i] = radius*Math.sin((2.0*Math.PI*i)/rows)+center[1];
  return YcircleCoordinates[i];
}

for (var  i = 0; i < 10; i++){
  $('p').prepend(xCord(i));
}

Answer 1

您的代码语法错误很少，下面应该有效。

＆＃13;

//screen size will use screen.width and screen.height later
var sW = 1920;
var sH = 1080;

//two arrays with the coordinates stored in them
var XcircleCoordinates = new Array();
var YcircleCoordinates = new Array();

//rows should be rows in database. I will use php to get the information, but assume 4 for now.
var rows = 4;

//radius and center of circle.
var radius = 200;
var center = [sW/2,sH/2];

function xCord(i){
   XcircleCoordinates[i] = radius*Math.cos((2.0*Math.PI*i)/rows)+center[0];
  return XcircleCoordinates[i];
}

function yCord(i){
  YcircleCoordinates[i] = radius*Math.sin((2.0*Math.PI*i)/rows)+center[1];
  return YcircleCoordinates[i];
}

for (var  i = 0; i < 10; i++){
  $('p').prepend(xCord(i));
}

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<p>Here is the coordinates: </p>

＆＃13;

使用JQuery迭代和打印坐标

1 个答案: