我正在尝试使用Rickshaw创建一些图表,并使用php中生成的ajax导入数据。
如果我使用静态数据,图表会显示;如果我复制/粘贴php中生成的数据(从控制台/log()),则会显示图表;如果我试图将数据放入一个变量中,并在js中使用该变量,它就是不起作用。:(
这是我从.php获得的控制台日志:(正如我所说,如果我将该代码块复制粘贴到.js中,并替换"dataOutEvo"var,则图形会显示出应有的效果。因此,我不认为数据是问题所在。
[
{
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()
}
]
这就是出错的地方:
$(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();
}
});
});
有人能告诉我出了什么问题吗?谢谢:)Mathieu
我现在在问自己,我是否不应该走另一条路,使用这个:
$fp = fopen('dataoutevo.json', 'w');
fwrite($fp, json_encode($js));
fclose($fp);
这个:
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 });
}
} );
但它仍然不起作用…有人能帮我告诉我做错了什么吗?
使用第一个实现应该可以正常工作。我认为你的问题是当你打电话时:
success: function(data) {
从PHP返回的数据变量实际上是一个字符串(您可以使用Javascript函数进行检查)-
console.log(typeof(data));
在您的PHP代码中,您应该返回一个(Association)Array,并确保您使用的是json_encode()函数-
echo json_encode($output);
在JS端,使用JSON.parse方法-投射返回的数据
var json_data = JSON.parse(data);
希望能有所帮助!
在您的第一个实现中,向ajax调用添加一个dataType。
这应该很好:
$(document).ready(function(){
$.ajax({
url: 'dataOutEvo.php', //le fichier qui va nous fournir la réponse
dataType: 'json',
success: function(data) {
var dataOutEvo = data;
console.log(dataOutEvo);
var palette = new Rickshaw.Color.Palette( { scheme: 'spectrum2001' } );
var graph = new Rickshaw.Graph({
element: document.querySelector("#chart"),
width: 960,
height: 260,
renderer: 'line',
series: dataOutEvo
});
graph.render();
}
});
});
请注意$.ajax调用中的dataType:'json'。
根据JQuery文档,添加json数据类型,
将响应求值为JSON并返回一个JavaScript对象。这个JSON数据以严格的方式进行解析;拒绝任何格式错误的JSON并且抛出解析错误