Javascript 使用下拉列表进行交叉筛选
我有两个下拉列表。 我希望每当在任一下拉列表中进行选择时,都会调用一个名为intersect的函数,该函数依次调用函数loadData和loadData1。我的代码附在这里Javascript 使用下拉列表进行交叉筛选,javascript,json,crossfilter,Javascript,Json,Crossfilter,我有两个下拉列表。 我希望每当在任一下拉列表中进行选择时,都会调用一个名为intersect的函数,该函数依次调用函数loadData和loadData1。我的代码附在这里 <select onchange="intersect()" id="metric" class="wrapper-dropdown"> <option value ="190">190</option> <option value ="90">90</optio
<select onchange="intersect()" id="metric" class="wrapper-dropdown">
<option value ="190">190</option>
<option value ="90">90</option>
<option value ="100">100</option>
<option value ="300">300</option>
</select>
<select onchange="intersect()" id="metric1" class="wrapper-dropdown">
<option value ="100">100</option>
<option value ="200">200</option>
<option value ="0">0</option>
</select>
<script>
var intersect= function() {
var total_90 = function() {
var data = [
{date: "2011-11-14T16:17:54Z", quantity: 2, total: 190, tip: 100, type: "tab"},
{date: "2011-11-14T16:20:19Z", quantity: 2, total: 190, tip: 100, type: "tab"},
{date: "2011-11-14T16:28:54Z", quantity: 1, total: 300, tip: 200, type: "visa"},
{date: "2011-11-14T16:30:43Z", quantity: 2, total: 90, tip: 0, type: "tab"},
{date: "2011-11-14T16:48:46Z", quantity: 2, total: 90, tip: 0, type: "tab"},
{date: "2011-11-14T16:53:41Z", quantity: 2, total: 90, tip: 0, type: "tab"},
{date: "2011-11-14T16:54:06Z", quantity: 1, total: 100, tip: 0, type: "cash"},
{date: "2011-11-14T16:58:03Z", quantity: 2, total: 90, tip: 0, type: "tab"},
{date: "2011-11-14T17:07:21Z", quantity: 2, total: 90, tip: 0, type: "tab"},
{date: "2011-11-14T17:22:59Z", quantity: 2, total: 90, tip: 0, type: "tab"},
{date: "2011-11-14T17:25:45Z", quantity: 2, total: 200, tip: 0, type: "cash"},
{date: "2011-11-14T17:29:52Z", quantity: 1, total: 200, tip: 100, type: "visa"}
];
var ndx = crossfilter(data);
var totalDim = ndx.dimension(function(d) { return d.total; });
var total_90 = totalDim.filter(document.getElementById('metric').selectedOptions[0].value);
return total_90;
};
var tip_90 = function() {
var data = [
{date: "2011-11-14T16:17:54Z", quantity: 2, total: 190, tip: 100, type: "tab"},
{date: "2011-11-14T16:20:19Z", quantity: 2, total: 190, tip: 100, type: "tab"},
{date: "2011-11-14T16:28:54Z", quantity: 1, total: 300, tip: 200, type: "visa"},
{date: "2011-11-14T16:30:43Z", quantity: 2, total: 90, tip: 0, type: "tab"},
{date: "2011-11-14T16:48:46Z", quantity: 2, total: 90, tip: 0, type: "tab"},
{date: "2011-11-14T16:53:41Z", quantity: 2, total: 90, tip: 0, type: "tab"},
{date: "2011-11-14T16:54:06Z", quantity: 1, total: 100, tip: 0, type: "cash"},
{date: "2011-11-14T16:58:03Z", quantity: 2, total: 90, tip: 0, type: "tab"},
{date: "2011-11-14T17:07:21Z", quantity: 2, total: 90, tip: 0, type: "tab"},
{date: "2011-11-14T17:22:59Z", quantity: 2, total: 90, tip: 0, type: "tab"},
{date: "2011-11-14T17:25:45Z", quantity: 2, total: 200, tip: 0, type: "cash"},
{date: "2011-11-14T17:29:52Z", quantity: 1, total: 200, tip: 100, type: "visa"}
];
var ndx = crossfilter(data);
var tipDim = ndx.dimension(function(d) { return d.tip; });
var tip_90 = tipDim.filter(document.getElementById('metric1').selectedOptions[0].value);
return tip_90;
}
var result = [];
for (var i = 0; i < total_90.length; i++){
if (total_90[i].total == 90 && total_90[i].tip == 0)
{
result.push(total_90[i]);
}
}
for (var i = 0; i < tip_90.length; i++){
if (tip_90[i].total == 90 && tip_90[i].tip == 0)
{
result.push(tip_90[i]);
}
}
function print_filter(filter){
var f=eval(filter);
if (typeof(f.length) != "undefined") {}else{}
if (typeof(f.top) != "undefined") {f=f.top(Infinity);}else{}
if (typeof(f.dimension) != "undefined") {f=f.dimension(function(d) { return "";}).top(Infinity);}else{}
console.log(filter+"("+f.length+") = "+JSON.stringify(f).replace("["," [\n\t").replace(/}\,/g,"},\n\t").replace("]","\n]"));
}
print_filter("result");
}
</script>
</body>
190
90
100
300
100
200
0
var intersect=function(){
var total_90=函数(){
风险值数据=[
{日期:“2011-11-14T16:17:54Z”,数量:2个,总数:190个,提示:100个,键入:“tab”},
{日期:“2011-11-14T16:20:19Z”,数量:2个,总数:190个,提示:100个,键入:“tab”},
{日期:“2011-11-14T16:28:54Z”,数量:1,总数:300,小费:200,类型:“visa”},
{日期:“2011-11-14T16:30:43Z”,数量:2,总数:90,提示:0,键入:“tab”},
{日期:“2011-11-14T16:48:46Z”,数量:2,总数:90,提示:0,键入:“tab”},
{日期:“2011-11-14T16:53:41Z”,数量:2,总数:90,提示:0,键入:“tab”},
{日期:“2011-11-14T16:54:06Z”,数量:1,总数:100,小费:0,键入:“现金”},
{日期:“2011-11-14T16:58:03Z”,数量:2,总数:90,提示:0,键入:“tab”},
{日期:“2011-11-14T17:07:21Z”,数量:2,总数:90,提示:0,键入:“tab”},
{日期:“2011-11-14T17:22:59Z”,数量:2,总数:90,提示:0,键入:“tab”},
{日期:“2011-11-14T17:25:45Z”,数量:2,总数:200,小费:0,类型:“现金”},
{日期:“2011-11-14T17:29:52Z”,数量:1,总数:200,小费:100,类型:“visa”}
];
var ndx=交叉滤波器(数据);
var totalDim=ndx.dimension(函数(d){返回d.total;});
var total_90=totalDim.filter(document.getElementById('metric')。selectedOptions[0]。value);
返回总数90;
};
var tip_90=函数(){
风险值数据=[
{日期:“2011-11-14T16:17:54Z”,数量:2个,总数:190个,提示:100个,键入:“tab”},
{日期:“2011-11-14T16:20:19Z”,数量:2个,总数:190个,提示:100个,键入:“tab”},
{日期:“2011-11-14T16:28:54Z”,数量:1,总数:300,小费:200,类型:“visa”},
{日期:“2011-11-14T16:30:43Z”,数量:2,总数:90,提示:0,键入:“tab”},
{日期:“2011-11-14T16:48:46Z”,数量:2,总数:90,提示:0,键入:“tab”},
{日期:“2011-11-14T16:53:41Z”,数量:2,总数:90,提示:0,键入:“tab”},
{日期:“2011-11-14T16:54:06Z”,数量:1,总数:100,小费:0,键入:“现金”},
{日期:“2011-11-14T16:58:03Z”,数量:2,总数:90,提示:0,键入:“tab”},
{日期:“2011-11-14T17:07:21Z”,数量:2,总数:90,提示:0,键入:“tab”},
{日期:“2011-11-14T17:22:59Z”,数量:2,总数:90,提示:0,键入:“tab”},
{日期:“2011-11-14T17:25:45Z”,数量:2,总数:200,小费:0,类型:“现金”},
{日期:“2011-11-14T17:29:52Z”,数量:1,总数:200,小费:100,类型:“visa”}
];
var ndx=交叉滤波器(数据);
var tipDim=ndx.dimension(函数(d){return d.tip;});
var tip_90=tipDim.filter(document.getElementById('metric1')。selectedOptions[0]。value);
返回提示_90;
}
var结果=[];
对于(变量i=0;i
因此,在任何给定的情况下,如果我在第一个下拉列表中选择“90”,并且在第二个下拉列表中已经选择了“0”,那么我希望在控制台上打印来自两个JSON数组(total=90和tip=0)的所有行。我不知道这里有什么困难 我希望无论何时在任一下拉列表中进行选择,都会调用一个名为intersect的函数,该函数依次调用函数loadData和loadData1
示例:我不确定您为什么要使用
d3.js
和crossfilter.js
我认为您的问题没有必要
因为您在问题中只标记了javascript
。我在此为您提供一个纯JS解决方案,而不是jquery
var m1 = document.getElementById('metric');
var m2 = document.getElementById('metric1');
m1.onchange = function(){
//alert(m1.value);
//alert(m2.value);
var result = intersection(data, data1, m1.value, m2.value);
if (result != ""){
for (var i = 0; i < result.length; i++){
console.log(result[i]);
}
}
}
m2.onchange = function(){
var result = intersection(data, data1, m1.value, m2.value);
if (result != ""){
for (var i = 0; i < result.length; i++){
console.log(result[i]);
}
}
}
//a is the data
//b is the data1
//v1 is the metric value
//v2 is the matric1 value
function intersection(a, b, v1, v2)
{
var result = [];
var c = a.concat(b);
for (var i = 0; i < c.length; i++){
if (c[i].total == v1 && c[i].tip == v2)
{
result.push(c[i]);
}
}
return result;
}
var m1=document.getElementById('metric');
var m2=document.getElementById('metric1');
m1.onchange=函数(){
//警报(m1.值);
//警报(m2.值);
var结果=交叉点(数据、数据1、m1.值、m2.值);
如果(结果!=“”){
对于(变量i=0;i
尝试选择total=90和tip=0。您将得到您想要的所谓“交叉点”结果。是的,很好。!!。我在这里也在做同样的事情。但是无论我做什么选择,我都会得到一个空白数组作为输出。如果你可以创建一些代码,请帮助我修复我发布的代码。把链接贴在这里,这样我可以更好地帮助你。但是,这段代码需要包含d3.js和crossfilter.js。我不知道如何在fiddle.netWhy上实现这一点。我需要d3和crossfilter是因为,我必须以图形和表格格式绘制值,当下拉列表中的任何一个值更改时,这些值都应该更改。事实上,还有很多下拉列表需要添加。我对所有这些编码和东西都是新手。
function loadData() {
alert("loadData reached");
}
function loadData1() {
alert("loadData1 reached");
}
function intersect() {
loadData();
loadData1();
}
var m1 = document.getElementById('metric');
var m2 = document.getElementById('metric1');
m1.onchange = function(){
//alert(m1.value);
//alert(m2.value);
var result = intersection(data, data1, m1.value, m2.value);
if (result != ""){
for (var i = 0; i < result.length; i++){
console.log(result[i]);
}
}
}
m2.onchange = function(){
var result = intersection(data, data1, m1.value, m2.value);
if (result != ""){
for (var i = 0; i < result.length; i++){
console.log(result[i]);
}
}
}
//a is the data
//b is the data1
//v1 is the metric value
//v2 is the matric1 value
function intersection(a, b, v1, v2)
{
var result = [];
var c = a.concat(b);
for (var i = 0; i < c.length; i++){
if (c[i].total == v1 && c[i].tip == v2)
{
result.push(c[i]);
}
}
return result;
}