gpt4 book ai didi

javascript - 在 JSON 中选择 Object.keys

转载 作者:行者123 更新时间:2023-11-30 14:50:41 25 4
gpt4 key购买 nike

想知道是否有人可以帮助我处理以下 JS?我正在尝试从 API 映射数据

var result = Object.keys(data).map(k => [new Date(k).getTime(), data[k]]);

这在像这样的数组上效果很好

{"2017-12-26":0,"2017-12-27":26,"2017-12-28":0}

但不是如果数组是

{"2017-12-21":{"nb_pageviews":10,"nb_uniq_pageviews":4,"nb_downloads":0,"nb_uniq_downloads":0,"nb_outlinks":4,"nb_uniq_outlinks":2,"nb_searches":0,"nb_keywords":0,"avg_time_generation":0.299},"2017-12-22":{"nb_pageviews":8,"nb_uniq_pageviews":4,"nb_downloads":0,"nb_uniq_downloads":0,"nb_outlinks":1,"nb_uniq_outlinks":1,"nb_searches":0,"nb_keywords":0,"avg_time_generation":0.453}}

我怎样才能只获取 nb_pageviews 的数据和每个日期的 value

这将在 NVD3 图表中用作

d3.json(`https://discovrbookings.innocraft.cloud/?module=API&method=VisitsSummary.getVisits&idSite=2&period=day&date=last30&format=json&token_auth=68aa5bd12137f13255dcb98794b65dff`, function(data1) {
d3.json(`https://discovrbookings.innocraft.cloud/?module=API&method=VisitsSummary.getUniqueVisitors&idSite=2&period=day&date=last30&format=json&token_auth=68aa5bd12137f13255dcb98794b65dff`, function(data2) {
d3.json(`https://discovrbookings.innocraft.cloud/?module=API&method=VisitsSummary.getActions&idSite=2&period=day&date=last30&format=json&token_auth=68aa5bd12137f13255dcb98794b65dff`, function(data3) {
d3.json(`https://discovrbookings.innocraft.cloud/?module=API&method=VisitsSummary.getBounceCount&idSite=2&period=day&date=last30&format=json&token_auth=68aa5bd12137f13255dcb98794b65dff`, function(data4) {
d3.json(`https://discovrbookings.innocraft.cloud/?module=API&method=API.get&format=json&idSite=2&period=day&date=last30&token_auth=68aa5bd12137f13255dcb98794b65dff`, function(data5) {

var result = Object.keys(data1).map(k => [new Date(k).getTime(), data1[k]]);
data1 = {
"key": "Visits",
"values": result
};
var result = Object.keys(data2).map(k => [new Date(k).getTime(), data2[k]]);
data2 = {
"key": "Unique Visits",
"values": result
};
var result = Object.keys(data3).map(k => [new Date(k).getTime(), data3[k]]);
data3 = {
"key": "Actions",
"values": result
};
var result = Object.keys(data4).map(k => [new Date(k).getTime(), data4[k]]);
data4 = {
"key": "Bounces",
"values": result
};
var result = Object.keys(data5)
.filter(k => !isNaN(data5[k].nb_pageviews))
.map(k => [new Date(k).getTime(), data5[k].nb_pageviews]);
data5 = {
"key": "Page Views",
"values": result
};

var data = [];
data.push(data1);
data.push(data2);
data.push(data3);
data.push(data4);
data.push(data5);


(function() {

nv.addGraph(function() {

var chart = nv.models.lineChart()
.margin({
left: 45,
right: 30
})
.x(function(d) {
return d[0]
})
.y(function(d) {
return d[1]
})

.useInteractiveGuideline(true);


chart.xAxis
.tickFormat(function(d) {
return d3.time.format('%m/%d/%y')(new Date(d))
});

chart.yAxis
.tickFormat(d3.format('d'));

chart.yAxis.tickPadding(15);
chart.xAxis.tickPadding(15);



d3.select('#summaryChart svg')
.datum(data)
.transition().duration(500)
.call(chart);

nv.utils.windowResize(chart.update);

$('#summaryChart').data('chart', chart);

return chart;
});

})();
});
});
});
});
});
#summaryChart {
height: 300px;
}
<link href="https://cdnjs.cloudflare.com/ajax/libs/nvd3/1.8.6/nv.d3.css" rel="stylesheet" />
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/3.4.11/d3.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/nvd3/1.8.6/nv.d3.js"></script>


<div id="summaryChart" class="line-chart text-center" data-x-grid="false">
<svg></svg>
</div>

最佳答案

只需更改访问 data[k].nb_pageviews 的方法即可

var result = Object.keys(data).map(k => [ new Date(k).getTime(), data[k].nb_pageviews ]);

有些日期没有针对它们的 nb_pageviews 属性,因此您可能还想将它们过滤掉

var result = Object.keys( data )
.filter( k => !isNaN( data[k].nb_pageviews ) )
.map(k => [ new Date(k).getTime(), data[k].nb_pageviews ] );

编辑

如果想要 '0' 而不是过滤掉那些没有 nb_pageviews 属性的日期,那么就制作它

data[k].nb_pageviews || 0

isNaN( data[k].nb_pageviews ) ? 0 : data[k].nb_pageviews || 0

最后

var result = Object.keys( data )
.map(k => [ new Date(k).getTime(), ( data[k].nb_pageviews || 0 ) ] );

关于javascript - 在 JSON 中选择 Object.keys,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48145647/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com