我正在尝试在力导向图上使用过滤器时遇到困难。我可以过滤掉节点,但不能使关联的链接消失。我对JavaScript的了解非常有限,但我猜逻辑应该是:如果节点被隐藏,则隐藏相关链接。我在正确的道路上吗? 如果有人可以帮助我,我将不胜感激!
数据格式:
{
"nodes": [
{"name":"AA1","group":"Group1","type":"a"},
{"name":"AA2","group":"Group2","type":"b"},
{"name":"AA3","group":"Group3","type":"c"},
{"name":"AA4","group":"Group4","type":"a"},
{"name":"AA5","group":"Group2","type":"b"},
{"name":"AA6","group":"Group4","type":"c"},...
],
"links": [
{"source":1,"target":59,"value":1},
{"source":1,"target":88,"value":1},
{"source":3,"target":12,"value":1},
{"source":3,"target":16,"value":1},
{"source":3,"target":87,"value":1},
{"source":5,"target":3,"value":1},
{"source":5,"target":16,"value":1},
{"source":5,"target":114,"value":1},... ]
过滤器代码:
// call method to create filter
createFilter();
// method to create filter
function createFilter()
{
d3.select(".filterContainer").selectAll("div")
.data(["a", "b", "c"])
.enter()
.append("div")
.attr("class", "checkbox-container")
.append("label")
.each(function(d) {
// create checkbox for each data
d3.select(this).append("input")
.attr("type", "checkbox")
.attr("id", function(d) {return "chk_" + d;})
.attr("checked", true)
.on("click", function(d, i) {
// register on click event
var lVisibility = this.checked? "visible":"hidden";
filterGraph(d, lVisibility);
})
d3.select(this).append("span")
.text(function(d){return d;});
});
$("#sidebar").show(); // show sidebar
}
// Method to filter graph
function filterGraph(aType, aVisibility)
{
// change the visibility of the node
node.style("visibility", function(o) {
var lOriginalVisibility = $(this).css("visibility");
return o.type === aType ? aVisibility : lOriginalVisibility;
});
/////////////////////////////////////// 需要代码隐藏链接 //////////////////////////////////////
}
答案 0 :(得分:1)
您需要通过检查其源或目标是否未被选中来隐藏链接。所以在你的filterGraph部分中,添加类似的东西(假设你的链接有class =“link”):
0x8180255
以Mike Bostock的悲惨为例,我使用上面的代码来过滤除了连接“Dahlia”和“Tholomyes”之外的所有其他代码。
以下是jsfiddle示例的片段:
positive = ["Dahlia", "Tholomyes"];
link.attr("display", function (o) {
////Here the structure of the the link can vary, sometimes it is o["source"]["name"], sometimes it is o["source"]["name"], check it out before you fill in.
var source_name = o["source"]["id"];
var target_name = o["target"]["id"];
var result = positive.indexOf(source_name) != -1 && positive.indexOf(target_name) != -1 ? "auto" : "none"
return result;
});