首页 \ 问答 \ dygraph 1.1.1似乎错误地检测到数据源的范围值(dygraph 1.1.1 appears to incorrectly detect range value for data source)

dygraph 1.1.1似乎错误地检测到数据源的范围值(dygraph 1.1.1 appears to incorrectly detect range value for data source)

我一直在努力想弄清楚如何让dygraphs与我的数据很好地配合。 当它渲染图表时,它会切断显示屏中的大部分线条。 但是,如果我向左侧滚动(移动+拖动),它会重新缩放视图并包含所有数据。 看起来图表末尾的下降正在扭曲图表的视图。

这很简单,我在这里有一个示例数据文件: http//pasamio.com/~pasamio/dygraph/sample_data.json

而这就是我用来吸引它的内容:

var dataURL = "http://pasamio.com/~pasamio/dygraph/sample_data.json";
var jsonData = null;

var jsonDataResult = $.ajax({
    url: dataURL,
    dataType: "json",
    async: false,
    success: (
        function(data) {
            jsonData = data;
        })
});

var data = new google.visualization.DataTable(jsonData);

var g = new Dygraph.GVizChart(document.getElementById("dg_div"));
g.draw(data, {"panEdgeFraction" : 0.1});

我在这里有一个示例jsfiddle,显示了坏的情况: https ://jsfiddle.net/g6b6jp9z/5/

任何想法都在这里发生了什么?


I've been struggling with trying to figure out how to get dygraphs to play nicely with my data. When it renders the chart, it cuts off a good chunk of the lines in the display. However if I scroll (shift + drag) to the left a little it rescales the view and includes all of the data. It looks like the drop off at the end of the graph is skewing the view of the chart.

It's really simple, I have a sample data file here: http://pasamio.com/~pasamio/dygraph/sample_data.json

And here's what I'm using to pull it in:

var dataURL = "http://pasamio.com/~pasamio/dygraph/sample_data.json";
var jsonData = null;

var jsonDataResult = $.ajax({
    url: dataURL,
    dataType: "json",
    async: false,
    success: (
        function(data) {
            jsonData = data;
        })
});

var data = new google.visualization.DataTable(jsonData);

var g = new Dygraph.GVizChart(document.getElementById("dg_div"));
g.draw(data, {"panEdgeFraction" : 0.1});

I've got a sample jsfiddle here that shows the bad case: https://jsfiddle.net/g6b6jp9z/5/

Any idea's what's going on here?


原文:https://stackoverflow.com/questions/41412788
更新时间:2023-03-03 15:03

最满意答案

$select = mysql_query("SELECT * FROM news WHERE timestamp <= NOW() ORDER BY timestamp");

试试......

请记住 - 您的帖子暗示您没有使用时间戳。 您似乎使用了datetime数据类型。 我的回答适用于日期时间。

另外,我认为你的头衔是误导性的。 根据您的问题,您似乎正在排队,所以事情将在您提供的日期之前发布。 如果在将来,那么它在该日期之前不会显示该文章。


$select = mysql_query("SELECT * FROM news WHERE timestamp <= NOW() ORDER BY timestamp");

try that...

Keep in mind - your post suggests you're not using a timestamp. You appear to be using a datetime datatype. My answer would work for datetime.

Also, I think your title is misleading. By your question it looks like you're making a queue so things will be posted by the date you provide. If in the future, then it does not display the article until that date.

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