我已经通读了几乎所有关于试图获取从PHP传递到javascript的数值的问题。正如你所看到的,我成功生成了随机数并且效果很好,但是当我试图将变量传递给graph.update函数时,即使两个变量的值都根据萤火虫(分别为14060116和0,这是正确的)他们没有把它更新到更新功能......任何想法?这是完整的脚本,如果它有帮助!变量不能从PHP传递到javascript
<script>
(function() {
function createCanvas(divName) {
var div = document.getElementById(divName);
var canvas = document.createElement('canvas');
div.appendChild(canvas);
if (typeof G_vmlCanvasManager != 'undefined') {
canvas = G_vmlCanvasManager.initElement(canvas);
}
var ctx = canvas.getContext("2d");
return ctx;
}
var ctx = createCanvas("graphDiv1");
var upld = <?php echo json_encode($dirupld[0]); ?>;
var upldred = <?php echo json_encode($dirupldred[0]); ?>;
var graph = new BarGraph(ctx);
graph.maxValue = 30;
graph.margin = 10;
graph.width = 450;
graph.height = 200;
graph.colors = ["#49a0d8", "#d353a0"];
graph.xAxisLabelArr = ["Traditional", "Duplicate Reduced"];
graph.update([upld, upldred]);
//graph.update([Math.random() * 30, Math.random() * 30]);
}());
</script>
不知道我失去了价值?萤火虫是报告以下
var upld = 14060116;
var upldred = 0;
graph.update([upld, upldred]);
做尝试把它们放在''“' – Deepak
如果你可以看到Firebug的变量,那不是纯粹的JavaScript问题吗? :) –
可以提供'json_encode($ dirupldred [0]);'从PHP脚本的示例输出?它看起来像'upld'获得一个整数,而不是JSON对象。 'upldred'也一样 –