Javascript HTML5画布线条未绘制
正如问题所暗示的,我正试图在HTML5画布上画一条线。我已经测试了这些值,发现它将在0,0到19201040(客户区的大小)之间画一条线,但对于输入的任何其他值都不会。我很确定这将是一个简单的错误,但我还没有发现它,我真的需要继续前进。谢谢你的帮助Javascript HTML5画布线条未绘制,javascript,html,canvas,Javascript,Html,Canvas,正如问题所暗示的,我正试图在HTML5画布上画一条线。我已经测试了这些值,发现它将在0,0到19201040(客户区的大小)之间画一条线,但对于输入的任何其他值都不会。我很确定这将是一个简单的错误,但我还没有发现它,我真的需要继续前进。谢谢你的帮助 function DrawGrid() { var canvas = document.getElementById("Grid"); var context = canvas.getContext("
function DrawGrid() {
var canvas = document.getElementById("Grid");
var context = canvas.getContext("2d");
context.beginPath();
context.moveTo(872, 432);
context.lineTo(1048, 432);
context.strokeStyle = "#FF0000";
context.lineWidth = 1;
context.stroke();
context.closePath();
}
代码如下:
function DrawGrid() {
var canvas = document.getElementById("Grid");
var context = canvas.getContext("2d");
context.beginPath();
context.moveTo(872, 432);
context.lineTo(1048, 432);
context.strokeStyle = "#FF0000";
context.lineWidth = 1;
context.stroke();
context.closePath();
}
您的
LineStartX
和LineStartY
变量一定有问题(无法测试,因为您没有提供CenterX
和CenterY
),用硬编码整数替换它们似乎可行
function DrawGrid() {
var canvas = document.getElementById("Grid");
var context = canvas.getContext("2d");
context.beginPath();
context.moveTo(872, 432);
context.lineTo(1048, 432);
context.strokeStyle = "#FF0000";
context.lineWidth = 1;
context.stroke();
context.closePath();
}
DrawGrid();
function DrawGrid() {
var canvas = document.getElementById("Grid");
var context = canvas.getContext("2d");
//var LineStartX = CenterX - (width / 2);
// var LineStartY = CenterY - (height / 2);
context.beginPath();
context.moveTo(10, 20);
context.lineTo(100, 20);
context.strokeStyle = "#FFAACC";
context.lineWidth = 1;
context.stroke();
}
我发现问题是由调用函数的位置引起的。在哪里定义了
CenterX
和CenterY
函数?在哪里调用函数?是什么导致了你的问题??
function DrawGrid() {
var canvas = document.getElementById("Grid");
var context = canvas.getContext("2d");
context.beginPath();
context.moveTo(872, 432);
context.lineTo(1048, 432);
context.strokeStyle = "#FF0000";
context.lineWidth = 1;
context.stroke();
context.closePath();
}