Javascript 类型错误:a.x未定义
我有一个像字典一样的列表数组Javascript 类型错误:a.x未定义,javascript,Javascript,我有一个像字典一样的列表数组 CONTORNO = [ {tipo: "m", x:[0,0]}, {tipo: "l", x:[0.06,0],x:[0.06,0.04],x:[0.14,0.04],x:[0.14,0],x:[0.24,0],x:[0.24,0.04],x:[0.34,0.04],x:[0.34,0], x:[0.44,0],x:[0.44,0.04],x:[0.54,0.04],x:[0.54,0],x:[0.64,0],x:[0.64,0.04],x:[0.74,0.04
CONTORNO = [
{tipo: "m", x:[0,0]},
{tipo: "l", x:[0.06,0],x:[0.06,0.04],x:[0.14,0.04],x:[0.14,0],x:[0.24,0],x:[0.24,0.04],x:[0.34,0.04],x:[0.34,0],
x:[0.44,0],x:[0.44,0.04],x:[0.54,0.04],x:[0.54,0],x:[0.64,0],x:[0.64,0.04],x:[0.74,0.04],x:[0.74,0],x:[0.8,0],
x:[0.8,1],x:[0.40,0.55],x:[0,1]}
]
然后是一个我调用这个数组的函数
function caminho(c,a)
{
c.beginPath();
for(var i=0;i<a.length;i++)
{
if(a[i].tipo=== "m")
{
c.moveTo(a.x[0],a.x[1]);
}else if (a[i].tipo === "q") {
c.quadraticCurveTo(a.x[0],a.x[1],a.x[2],a.x[3]);
}else if(a[i].tipo === "l") {
c.lineTo(a.x[0],a.x[1]);
}
}
c.closePath();
}
caminho(c,CONTORNO);
功能卡米尼奥(c,a)
{
c、 beginPath();
对于(var i=0;i对象必须具有属性x
然后,当访问循环中的对象时,应该说“i select”a“zero”a[0]并访问“x”属性
例如:
a[0].x
只需将c.moveTo(a.x[0],a.x[1]);
更改为c.moveTo(a[i].x[0],a[i].x[1]);
等等
功能卡米尼奥(c,a){
c、 beginPath();
对于(变量i=0;i caminho(c,CONTORNO);
使用a[1].x,因为a[1]是循环中的项