获取MySql数据并将其存储到Javascript数组中 [英] Get MySql data and store it into Javascript array

问题描述

想要从Mysql检索记录数据并将其存储为以下格式的heatmap.js地图数据的Javascript数组:

want to retrieve a records data from Mysql and store it into Javascript array for heatmap.js map data in this format :

var testData = {max: 46, data: [{lat: 33.5363, lon:-117.044, value: 1},{lat: 33.5608, lon:-117.24, value: 1},..]};

现在我被困在这里，而且我不知道如何从Jquery连接到我的var testData = new Array();，我该怎么做才能解决这个问题?

Now I get stuck at here, and I don't know how to connecting from Jquery into my var testData = new Array();, How I should do to solve this?

(更新后的正确代码)

get_query.php

get_query.php

<?php
require_once('./db_con.php');
$dbcon=new db;


$query="SELECT (SELECT geo_lat FROM fun WHERE created_at <= DATE_SUB(NOW(), interval 1   minute) AS geo_lat," . 
       "(SELECT geo_long FROM fun WHERE created_at <= DATE_SUB(NOW(), interval 1 minute) AS geo_long";

$result = mysqli_query($dbcon,$query);
$data = array(); 

    while($row= mysqli_fetch_assoc($result)){

     $data[] = array("lat"=>$row["geo_lat"], "lon"=>$row["geo_long"], "value"=>1); 
     $post_data = json_encode(array('max' => 46, 'data' => $data));
    }
    echo $post_data;
 ?>

my_data.js来自此处:

my_data.js based from here:

jQuery.extend({
getValues: function(url) {
    var result = null;
    $.ajax({
        url: url,
        type: 'get',
        dataType: 'json',
        async: false,
        success: function(data) {
            result = JSON.stringify(data);
        }
    });
   return result;
}
});

var testData = $.getValues("get_query.php");

感谢Orangepill和Chrislondon.

Thanks to Orangepill and Chrislondon.

推荐答案

所以您的问题是如何将jQuery连接到var data，这样我就不会在PHP代码中遇到很多问题.在成功功能中，您可以这样设置var data:

So your question is how to connect your jQuery to your var data so I won't get into the myriad of problems in your PHP code. In your success function you can set the var data like so:

var data = new Array();

$(function() {
    $.ajax({
        type:     "post",
        url:      "get_query.php",
        data:     $(this).serialize(),
        dataType: "json"
}).done(function(response) {
    data = response;
});

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