我正在尝试创建一个包含内联geojson对象中列出的每个功能的属性的表。
以下几乎得到我想要的但它只显示第一个条目的属性。我应该如何更改它以显示所有条目的属性?
var thedatatable = d3.select("#propertiestable")
.append("table")
.attr("class", "datatable");
var header = thedatatable.selectAll("th")
.data(d3.keys(datapointsjson.features[0].properties))
.enter()
.append("th")
.text(function (d) {
return d
});
var tr = thedatatable.selectAll("tr")
.data(d3.values(datapointsjson.features[0].properties))
.enter()
.append("tr");
var td = thedatatable.selectAll("td")
.data(d3.values(datapointsjson.features[0].properties))
.enter()
.append("td")
.text(function (d) {
return d
});
我有理由相信这是“td”位我错了,但我尝试的不仅仅是第一个条目。
这是我正在使用的geojson,它需要在同一个文件中内联。
var datapointsjson = {
"type": "FeatureCollection",
"features": [{
"type": "Feature",
"geometry": {
"type": "Point",
"coordinates": [
136.33, -31.5
]
},
"properties": {
"regno": "R123456",
"taxon": "genus1 species1",
"sitecode": "",
"nearestnamedplace": "FREELING ISLAND",
"preciselocation": "south coast",
"collection": "Herpetology"
}
}, {
"type": "Feature",
"geometry": {
"type": "Point",
"coordinates": [
137.07, -36.23
]
},
"properties": {
"regno": "R654185",
"taxon": "genus2 species2",
"sitecode": "",
"nearestnamedplace": "Neptune Island",
"preciselocation": "Middle of island",
"collection": "Herpetology"
}
}, {
"type": "Feature",
"geometry": {
"type": "Point",
"coordinates": [
142.0358, -38.7719
]
},
"properties": {
"regno": "R6528445",
"taxon": "genus3 species3",
"sitecode": "TT02",
"nearestnamedplace": "Woolongong",
"preciselocation": "5 km N",
"collection": "Herpetology"
}
}, {
"type": "Feature",
"geometry": {
"type": "Point",
"coordinates": [
137.6914, -32.2789
]
},
"properties": {
"regno": "R654987",
"taxon": "genus4 species4",
"sitecode": "IL0601",
"nearestnamedplace": "Ballarat",
"preciselocation": "5.3 KM E",
"collection": "Herpetology"
}
}]
};
我做了jsfiddle显示结果
任何建议都会感激不尽。谢谢。
答案 0 :(得分:0)
我已经把你的小提琴碰到了here很多基于Shawn Allen的精彩回答here。
现在,您需要提供一个数组,以便d3可以迭代它们(在这种情况下为4次)
var tr = thedatatable.selectAll("tr")
.data(d3.values(datapointsjson.features))
.enter()
.append("tr");
然后你需要使用以下方法从json中提取所需的信息:
var td = tr.selectAll("td")
.data(function(row) {
return headers.map(function(d) { console.log(row.properties)
return {column: d, value: row.properties[d]};
});
})
.enter()
.append("td")
.text(function (d) {
return d.value;
});