一页中有多个ajax请求+怎么做+最佳实践 [英] multiple ajax requests from one page + how to do + best practice

问题描述

我目前正在研究Google图表，并且已经完成了一个基本的设置.

I am working on google charts at the minute and I have got a basic one setup.

它目前所做的是连接到DB，并基于1个查询返回一个数据集.我想知道的是，如果我想向数据库绘制更多具有不同查询的图表，该怎么做?或最佳做法是什么?

What it does at present is connects to a DB and returns a dataset based on 1 query. I am wondering is, if I want to draw more charts with different queries to the database how do I do this? Or what is the best practice?

例如，一个查询已经存在一个连接，如何添加另一个查询，然后根据返回的内容绘制图表?

For instance, there is already one connection with one query, how can I add another query, and then draw the charts based on what is returned?

我知道这可能是一个广泛的问题，但是也许有人可以向我展示如何从数据库返回不同的查询/数据集?

I understand that might be a broad question, but maybe someone could show me how I would return a different query/dataset from the DB?

这是我的代码:

 $(document).ready(function(){

    console.log("hello world")
    //alert("result")

    $.ajax({
        url:"data.php",
        dataType : "json",
        success : function(result) {
            google.charts.load('current', {
                'packages':['corechart','bar']
            });
            google.charts.setOnLoadCallback(function() {
                console.log(result[0]["name"])
                drawChart(result);                  
            });
        }
    }); 

    //add a 2nd call - will need a 2nd draw charts to draw the different dataset assuming it will be different
    //                  - will need a 2nd data.php as the query will be different on the dataset
    $.ajax({
        url:"data.php",
        dataType : "json",
        success : function(result2) {
            google.charts.load('current', {
                'packages':['corechart','bar']
            });
            google.charts.setOnLoadCallback(function() {
                console.log(result2[0]["name"])
                drawChart(result2);                 
            });
        }
    }); 

    function drawChart(result) {

        var data = new google.visualization.DataTable();
        data.addColumn('string','Name');
        data.addColumn('number','Quantity');
        var dataArray=[];
        $.each(result, function(i, obj) {
            dataArray.push([ obj.name, parseInt(obj.quantity) ]);
        });

        data.addRows(dataArray);

        var piechart_options = {
            title : 'Pie Chart: How Much Products Sold By Last Night',
            width : 400,
            height : 300
        }
        var piechart = new google.visualization.PieChart(document
                .getElementById('piechart_div'));
        piechart.draw(data, piechart_options)

        var columnchart_options = {
            title : 'Bar Chart: How Much Products Sold By Last Night',
            width : 400,
            height : 300,
            legend : 'none'
        }
        //var barchart = new google.visualization.BarChart(document
            //  .getElementById('barchart_div'));               
        //barchart.draw(data, barchart_options)

        var chart = new google.charts.Bar(document.getElementById('columnchart_material'));
        chart.draw(data, google.charts.Bar.convertOptions(columnchart_options));
    }           //have added this column chart but need to wrok out if it is best practice????

}); 

我从数据库查询中得到一个对象，但是我想知道如何从同一数据库连接返回更多/不同的数据集?例如，如果我想用此查询select * from product where name="Product1" OR name="Product2";

I am getting an object back from my DB query, but I want to know how to return more/different datasets from the same DB connection? For example what if I wanted to draw another chart with the dataset returned from this query select * from product where name="Product1" OR name="Product2";

0: Object { id: "1", name: "Product1", quantity: "2" }
​1: Object { id: "2", name: "Product2", quantity: "3" }
​2: Object { id: "3", name: "Product3", quantity: "4" }
​3: Object { id: "4", name: "Product4", quantity: "2" }
​4: Object { id: "5", name: "Product5", quantity: "6" }
​5: Object { id: "6", name: "Product6", quantity: "11" }

对于我的PHP代码而言，如下所示:

For what it is worth my php code is as follows:

data.php

<?php
require_once 'database.php';
$stmt = $conn->prepare('select * from product');
$stmt->execute();
$results = $stmt->fetchAll(PDO::FETCH_OBJ);
echo json_encode($results); 
?>

database.php

database.php

<?php
$conn = new PDO('mysql:host=192.168.99.100;dbname=demo','root', 'root');
?>

注意:这可能很有趣

推荐答案

每次加载一次Google图表只需加载一次，
并非每次您都需要绘制图表

google charts only needs to be loaded once per page load,
not every time you need to draw a chart

也可以使用google.charts.load代替-> $(document).ready
它将在执行回调/Promise

also, google.charts.load can be used in place of --> $(document).ready
it will wait for the page to load before executing the callback / promise

推荐类似于以下代码段的设置...

recommend setup similar to following snippet...

google.charts.load('current', {
  packages: ['corechart', 'bar']
}).then(function () {
  $.ajax({
    url: 'data.php',
    dataType: 'json'
  }).done(drawChart1);

  $.ajax({
    url: 'data.php',
    dataType: 'json'
  }).done(drawChart2);
});

function drawChart1(result) {
  ...
}

function drawChart2(result) {
  ...
}

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