我试图用Rickshaw创建一些图表,用php中生成的ajax导入数据。人力车:多个系列不能正常工作的数据
如果我使用静态数据,图表显示; 如果我复制/粘贴在php中生成的数据(从控制台/日志()),图表显示; 如果我试图将数据放入一个变量中,并在js中使用该变量,它就不起作用。 :(
这是我从.php为得到我的控制台日志: (就像我说的,如果我复制粘贴代码块到.js文件替换“dataOutEvo”变种,该图显示,因为它。应该因此,我不认为数据是问题
[
{
name: "ligne",
data: [{x:0,y:35},{x:1,y:34},{x:2,y:36},{x:3,y:35},{x:4,y:40},{x:5,y:35},{x:6,y:37},{x:7,y:40},{x:8,y:45},{x:9,y:46},{x:10,y:55},{x:11,y:63},{x:12,y:61},{x:13,y:45},{x:14,y:48},{x:15,y:49},{x:16,y:45},{x:17,y:44},{x:18,y:52},{x:19,y:43},{x:20,y:37},{x:21,y:36},{x:22,y:37},{x:23,y:34}],
color: palette.color()
},
{
name: "ligne",
data: [{x:0,y:10},{x:1,y:15},{x:2,y:13},{x:3,y:15},{x:4,y:14},{x:5,y:16},{x:6,y:17},{x:7,y:25},{x:8,y:23},{x:9,y:24},{x:10,y:25},{x:11,y:28},{x:12,y:27},{x:13,y:21},{x:14,y:23},{x:15,y:19},{x:16,y:18},{x:17,y:16},{x:18,y:15},{x:19,y:14},{x:20,y:15},{x:21,y:16},{x:22,y:15},{x:23,y:16}],
color: palette.color()
},
{
name: "ligne",
data: [{x:0,y:45},{x:1,y:49},{x:2,y:49},{x:3,y:50},{x:4,y:54},{x:5,y:51},{x:6,y:54},{x:7,y:65},{x:8,y:68},{x:9,y:70},{x:10,y:80},{x:11,y:91},{x:12,y:88},{x:13,y:66},{x:14,y:71},{x:15,y:68},{x:16,y:63},{x:17,y:60},{x:18,y:67},{x:19,y:57},{x:20,y:52},{x:21,y:52},{x:22,y:52},{x:23,y:50}],
color: palette.color()
},
{
name: "ligne",
data: [{x:0,y:10},{x:1,y:15},{x:2,y:12},{x:3,y:5},{x:4,y:9},{x:5,y:15},{x:6,y:45},{x:7,y:125},{x:8,y:345},{x:9,y:256},{x:10,y:312},{x:11,y:345},{x:12,y:299},{x:13,y:165},{x:14,y:354},{x:15,y:368},{x:16,y:254},{x:17,y:213},{x:18,y:312},{x:19,y:165},{x:20,y:54},{x:21,y:32},{x:22,y:10},{x:23,y:5}],
color: palette.color()
},
{
name: "ligne",
data: [{x:0,y:2},{x:1,y:3},{x:2,y:2},{x:3,y:1},{x:4,y:1},{x:5,y:2},{x:6,y:3},{x:7,y:15},{x:8,y:45},{x:9,y:27},{x:10,y:40},{x:11,y:42},{x:12,y:35},{x:13,y:18},{x:14,y:42},{x:15,y:40},{x:16,y:30},{x:17,y:25},{x:18,y:40},{x:19,y:20},{x:20,y:6},{x:21,y:4},{x:22,y:2},{x:23,y:1}],
color: palette.color()
}
]
而这正是出错的JS:
$(document).ready(function(){
$.ajax({
url: 'dataOutEvo.php', //le fichier qui va nous fournir la réponse
success: function(data) {
var dataOutEvo = data;
console.log(dataOutEvo);
var palette = new Rickshaw.Color.Palette({ scheme: 'spectrum2001' });
var graph = new Rickshaw.Graph({
element: document.querySelector("#chart"),
width: 960,
height: 260,
renderer: 'line',
series: dataOutEvo
});
graph.render();
}
});
});
有人能告诉我什么不顺心吗?谢谢: )Mathieu
我问自己,如果现在我不应该走另一条路,使用此:
$fp = fopen('dataoutevo.json', 'w');
fwrite($fp, json_encode($js));
fclose($fp);
这:
var palette = new Rickshaw.Color.Palette();
new Rickshaw.Graph.Ajax({
element: document.getElementById("chart"),
width: 800,
height: 500,
renderer: 'line',
dataURL: 'dataoutevo.json',
onData: function(d) {
Rickshaw.Series.zeroFill(d);
return d;
},
onComplete: function(transport) {
var graph = transport.graph;
var detail = new Rickshaw.Graph.HoverDetail({ graph: graph });
}
});
但它仍然没有工作...有谁可以帮助我,告诉我我做错了什么?
你能否解释多一点,你是什么意思时,你说 - _IF我尝试把DATAS到一个变量,并使用该变种在JS ,它只是不” WORK_ ,然后当你继续说 _if我复制的代码块粘贴到.js文件替换‘dataOutEvo’变种,该图显示,因为它应该)_ ? 我刚刚[** fiddled **](http://jsfiddle.net/srvikram13/Ar6BS/)你的代码与硬编码的数据,它似乎工作正常。 –