Question

我正在尝试抓取光标所在的最近绘制点。我

我在findNearbyItem源代码中找到了似乎能够执行此操作的jquery.flot.js函数，但是当我手动调用它时，我收到了ReferenceError: findNearbyItem is not defined错误。

这是我所指的功能：

function findNearbyItem(mouseX, mouseY, seriesFilter) {
    var maxDistance = options.grid.mouseActiveRadius,
        smallestDistance = maxDistance * maxDistance + 1,
        item = null, foundPoint = false, i, j, ps;

    for (i = series.length - 1; i >= 0; --i) {
        if (!seriesFilter(series[i]))
            continue;

        var s = series[i],
            axisx = s.xaxis,
            axisy = s.yaxis,
            points = s.datapoints.points,
            mx = axisx.c2p(mouseX), // precompute some stuff to make the loop faster
            my = axisy.c2p(mouseY),
            maxx = maxDistance / axisx.scale,
            maxy = maxDistance / axisy.scale;

        ps = s.datapoints.pointsize;
        // with inverse transforms, we can't use the maxx/maxy
        // optimization, sadly
        if (axisx.options.inverseTransform)
            maxx = Number.MAX_VALUE;
        if (axisy.options.inverseTransform)
            maxy = Number.MAX_VALUE;

        if (s.lines.show || s.points.show) {
            for (j = 0; j < points.length; j += ps) {
                var x = points[j], y = points[j + 1];
                if (x == null)
                    continue;

                // For points and lines, the cursor must be within a
                // certain distance to the data point
                if (x - mx > maxx || x - mx < -maxx ||
                    y - my > maxy || y - my < -maxy)
                    continue;

                // We have to calculate distances in pixels, not in
                // data units, because the scales of the axes may be different
                var dx = Math.abs(axisx.p2c(x) - mouseX),
                    dy = Math.abs(axisy.p2c(y) - mouseY),
                    dist = dx * dx + dy * dy; // we save the sqrt

                // use <= to ensure last point takes precedence
                // (last generally means on top of)
                if (dist < smallestDistance) {
                    smallestDistance = dist;
                    item = [i, j / ps];
                }
            }
        }

        if (s.bars.show && !item) { // no other point can be nearby
            var barLeft = s.bars.align == "left" ? 0 : -s.bars.barWidth/2,
                barRight = barLeft + s.bars.barWidth;

            for (j = 0; j < points.length; j += ps) {
                var x = points[j], y = points[j + 1], b = points[j + 2];
                if (x == null)
                    continue;

                // for a bar graph, the cursor must be inside the bar
                if (series[i].bars.horizontal ?
                    (mx <= Math.max(b, x) && mx >= Math.min(b, x) &&
                     my >= y + barLeft && my <= y + barRight) :
                    (mx >= x + barLeft && mx <= x + barRight &&
                     my >= Math.min(b, y) && my <= Math.max(b, y)))
                        item = [i, j / ps];
            }
        }
    }

    if (item) {
        i = item[0];
        j = item[1];
        ps = series[i].datapoints.pointsize;

        return { datapoint: series[i].datapoints.points.slice(j * ps, (j + 1) * ps),
                 dataIndex: j,
                 series: series[i],
                 seriesIndex: i };
    }

    return null;
}

如果有其他方法可以解决此问题，请告知我们。

Answer 1

你的用例是什么？如果将mouseActiveRadius选项增加到大型flot，将为您找到最接近光标的位置。

var options = {
    grid: { 
        hoverable: true, 
        mouseActiveRadius: 1000
    }
}

示例here。

修改

是的，您可以使用plothover事件来检索突出显示的点。

$("#placeholder").bind("plothover", function (event, pos, item) { if (item){ var x = item.datapoint[0].toFixed(2), y = item.datapoint[1].toFixed(2); console.log("x:" + x + ", " + "y:" + y); } });

更新了小提琴here。

Answer 2

这是flot的内部功能，但您可以自己轻松地重新创建相同的功能。您需要做的就是遍历数据系列中的点并将它们的位置与鼠标指针进行比较（毕达哥拉斯应该能够在这里帮助您）。

Answer 3

至于您看到ReferenceError: findNearbyItem is not defined的原因，请参阅this。手动应用这个小补丁将使功能公开。

Flot：如何获取光标所在的最近绘制点？

3 个答案: