Javascript 更改函数时JSXgraph不移动点
我用JSXgraph向学生展示LIMSUP和liminfs。这个版本在这里起作用- 代码如下:Javascript 更改函数时JSXgraph不移动点,javascript,jsxgraph,Javascript,Jsxgraph,我用JSXgraph向学生展示LIMSUP和liminfs。这个版本在这里起作用- 代码如下: <div style="width: 500px; height: 40px;"> <p style="display: inline;">N:</p> // Create a slider for the values of N <input id="NSlide
<div style="width: 500px; height: 40px;">
<p style="display: inline;">N:</p>
// Create a slider for the values of N
<input id="NSlider" type="range" min="1" max="19" value="1" step="1" style="width: 50%;">
<p style="display: inline;"><span id="NOut"></span></p>
<script>
// Get the value of N from the slider
var NSlider = document.getElementById("NSlider");
var NOutput = document.getElementById("NOut");
NOutput.innerHTML = NSlider.value; // Output value for student to see
</script>
</div>
<!-- Create board with points -->
<div id="box" class="jxgbox" style="width:500px; height:500px;">
<script type="text/javascript">
// Create board
var board = JXG.JSXGraph.initBoard('box', {
boundingbox: [-1, 2, 21, -0.2],
axis: true,
grid: true
});
// Generate points in the sequence and graph points of the sequence
var i;
var s = [null];
for (i = 1; i <= 20; i++) {
s.push(1 + Math.pow(-1, i) / i);
board.create('point', [i, s[i]], {
color: 'yellow',
fixed: true,
withLabel: false
});
}
// Genereate liminfs and limsups
var infs = [null],
sups = [null];
for (i = 1; i <= 20; i++) {
infs.push(Math.min.apply(null, s.slice(i + 1)));
sups.push(Math.max.apply(null, s.slice(i + 1)));
}
// Graph liminf and limsup points
var liminf = board.create('point', [
function() { return NSlider.value; },
function() { return infs[NSlider.value]; }], {
color: 'blue',
withLabel: false
});
var limsup = board.create('point', [
function() { return NSlider.value; },
function() { return sups[NSlider.value]; }], {
color: 'orange',
withLabel: false
});
board.update()
</script>
</div>
<script>
// Set board to update when the N slider is updated
NSlider.oninput = function () {
NOutput.innerHTML = this.value; // Output value for student to see
liminf.moveTo([NSlider.value, infs[NSlider.value]]);
limsup.moveTo([NSlider.value, sups[NSlider.value]]);
}
</script>
N:
//为N的值创建一个滑块
//从滑块获取N的值
var NSlider=document.getElementById(“NSlider”);
var NOutput=document.getElementById(“NOut”);
NOutput.innerHTML=NSlider.value;//输出值供学生查看
//创建板
var board=JXG.JSXGraph.initBoard('box'{
边界框:[-1,2,21,-0.2],
安讯士:没错,
网格:对
});
//在序列中生成点并绘制序列的点
var i;
var s=[null];
对于(i=1;iu
和v
在您的绘图中是什么意思?我们如何计算它?在网站上的文本中,u\N是数组infs的元素,v\N是数组sups的元素DOM上有蓝点,但一些脚本正在设置显示:无;可见性:隐藏;
样式。您必须取消错误。我在电路板上的点上添加了visible:true
,但它没有解决问题。知道当我更改填充s的函数时为什么会这样做吗?我已经确定了问题。如果sups或infs中的值为负值,moveTo函数不会显示该点。