我正在使用jqplot并且我绘制了一个来自数组php的数据但是当我尝试从javascript绘图时不能正常工作
这是代码php:
$array= array();
$array[] = 34.890;
$array[] = 25.090;
现在代码吸收了:
<script>
$(function() {
var array = new Array();
<?php
for( $i=0; $i<count($array); $i++){
echo "\narray[$i] = '$array[$i]';";
}
?>
graficar( array );
});
</script>
graficar函数是:
function graficar( array )
{
var plot2 = $.jqplot('lienzo',
[
array
],
{
seriesDefaults: {
renderer: $.jqplot.BarRenderer,
// Show point labels to the right ('e'ast) of each bar.
// edgeTolerance of -15 allows labels flow outside the grid
// up to 15 pixels. If they flow out more than that, they
// will be hidden.
pointLabels: {show: true, location: 'e', edgeTolerance: -15},
// Rotate the bar shadow as if bar is lit from top right.
shadowAngle: 135,
// Here's where we tell the chart it is oriented horizontally.
rendererOptions: {
barDirection: 'horizontal',
barWidth:18
}
},
axes: {
yaxis: {
renderer: $.jqplot.CategoryAxisRenderer
}
}
}
);
}
我得到的错误是:
TypeError: this[r]._ticks[0] is undefined
[Break On This Error]
...{href:function(a){return a.getAttribute("href")},type:function(a){return a.getAt...
图表显示:
如何解决我的麻烦?
坦克。
答案 0 :(得分:0)
你的&#34;数组&#34;中出现了问题,就像在我的代码中一样,
array = ["0","0","0","0"]
或
array = ["w","r","o","n","g"]
发生错误,然后我应该使用parseInt方法将字符串元素解析为第一个整数。如果
array = ["1","2","3","4"]
没有必要将字符串元素解析为数组中的intger,jqplot会为你做,但如果数组就像前两个一样,这里就会出现错误。