如何从数据库中获取值并在图表或图形中显示它 [英] How Can I Fetch The Value From Databse And Display It In Chart Or Graph

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本文介绍了如何从数据库中获取值并在图表或图形中显示它的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

  $ query  =   SELECT name FROM mybase WHERE name ='$ name'; 

$ mysqli = new mysqli();
$ mysqli-> connect(' localhost'' blabla'' blabla'' blabla');
$ result = $ mysqli-> query($ query);
while ($ row = $ result-> fetch_assoc()){
// 这里我无法弄清楚如何以饼图的格式制作数据
}


< html >
< span class =code-keyword>< head >
< script 类型 = text / javascript src = https ://www.google.com/jsapi > < / script >
< script 类型 = text / javascript >
google.load( 可视化 1,{packages:[ corechart]});
google.setOnLoadCallback(drawChart);
function drawChart(){
var data = google.visualization.arrayToDataTable( [
[' 名称'' 看x次'],
[' somenamefrommysql' 11 ],
[' somenamefrommysql' 2 ],
[' somenamefrommysql' 2 ],
[ ' somenamefrommysql' 2 ],
[' somenamefrommysql' 7 ]
]);

var options = {
title:' 我的每日活动
};

var chart = new google.visualization.PieChart( document .getElementById(' chart_div'));
chart.draw(数据,选项);
}
< / 脚本 >
< / head >
< body >
< div id = chart_div style < span class =code-keyword> = width:900px; height:500px; > < / div >
< / body >
< / html >

解决方案

query = SELECT name FROM mybase WHERE name ='


name';


mysqli = new mysqli的();

$query = "SELECT name FROM mybase WHERE name = '$name' ";

        $mysqli = new mysqli();
        $mysqli->connect('localhost', 'blabla', 'blabla', 'blabla');
        $result = $mysqli->query($query);
        while ($row = $result->fetch_assoc()) {
            //here I cant figure out how to make data in format for pie chart
        }


<html>
  <head>
    <script type="text/javascript" src="https://www.google.com/jsapi"></script>
    <script type="text/javascript">
      google.load("visualization", "1", {packages:["corechart"]});
      google.setOnLoadCallback(drawChart);
      function drawChart() {
        var data = google.visualization.arrayToDataTable([
          ['Name', 'Seen x times'],
          ['somenamefrommysql',     11],
          ['somenamefrommysql',      2],
          ['somenamefrommysql',  2],
          ['somenamefrommysql', 2],
          ['somenamefrommysql',    7]
        ]);

        var options = {
          title: 'My Daily Activities'
        };

        var chart = new google.visualization.PieChart(document.getElementById('chart_div'));
        chart.draw(data, options);
      }
    </script>
  </head>
  <body>
    <div id="chart_div" style="width: 900px; height: 500px;"></div>
  </body>
</html>

解决方案

query = "SELECT name FROM mybase WHERE name = '


name' ";


mysqli = new mysqli();


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