Question

我想创建一个饼图，它基本上应该显示学生已经完成的所有任务中完成了多少任务。所以我的任务表看起来像这样

任务 task_id | task_student_id | task_status

* task_student_id *引用学生表中的* user_id *

因此，假设我的带有user_id的学生的样本数据在任务表

中如下所示

task_id | task_student_id | task_status 1 6完成 2 6完成 3 6 not_complete 4 6 not_complete 5 6完成

因此，该学生的饼图应显示已完成和未完成任务的数量或百分比（即40％未完成60％完成）

我在网上找到了一个教程。有两个php文件，第一个允许你从学生表中选择user_id

<html>
<head>
  <!--Load the AJAX API-->
  <script type="text/javascript" src="http://www.google.com/jsapi"></script>
  <script type="text/javascript" src="jquery-1.7.1.min.js"></script>
  <script type="text/javascript">

  // Load the Visualization API and the piechart,table package.
  google.load('visualization', '1', {'packages':['corechart','table']});

  function drawItems(num) {
    var jsonPieChartData = $.ajax({
      url: "getpiechartdata.php",
      data: "q="+num,
      dataType:"json",
      async: false
    }).responseText;

    var jsonTableData = $.ajax({
      url: "gettabledata.php",
      data: "q="+num,
      dataType:"json",
      async: false
    }).responseText;

    // Create our data table out of JSON data loaded from server.
    var piechartdata = new google.visualization.DataTable(jsonPieChartData);
    var tabledata = new google.visualization.DataTable(jsonTableData);

    // Instantiate and draw our pie chart, passing in some options.
    var chart = new google.visualization.PieChart(document.getElementById('chart_div'));
    chart.draw(piechartdata, {
      width: 700,
      height: 500,
      chartArea: { left:"5%",top:"5%",width:"90%",height:"90%" }
    });

    // Instantiate and draw our table, passing in some options.
    var table = new google.visualization.Table(document.getElementById('table_div'));
    table.draw(tabledata, {showRowNumber: true, alternatingRowStyle: true});
  }

  </script>
</head>
<body>
  <form>
  <select name="users" onchange="drawItems(this.value)">
  <option value="">Select a student:</option>
  <?php
    $dbuser="";
    $dbname="";
    $dbpass="";
    $dbserver="";
    // Make a MySQL Connection
    mysql_connect($dbserver, $dbuser, $dbpass) or die(mysql_error());
    mysql_select_db($dbname) or die(mysql_error());
    // Create a Query
    $sql_query = "SELECT user_id, user_name FROM students";
    // Execute query
    $result = mysql_query($sql_query) or die(mysql_error());
    while ($row = mysql_fetch_array($result)){
    echo '<option value='. $row['user_id'] . '>'. $row['user_name'] . '</option>';
    }
    mysql_close($con);
  ?>
  </select>
  </form>
  <div id="chart_div"></div>
  <div id="table_div"></div>
</body>
</html>

然后第二个php文件根据所选的user_id

填充图表

<?php
  $q=$_GET["q"];

  $dbuser="";
  $dbname="";
  $dbpass="";
  $dbserver="";

  $sql_query = "SELECT task_status, COUNT(*) FROM tasks
    WHERE  task_student_id=" . $q . ""

  $con = mysql_connect($dbserver,$dbuser,$dbpass);
  if (!$con){ die('Could not connect: ' . mysql_error()); }
  mysql_select_db($dbname, $con);

  $result = mysql_query($sql_query);

    $data = array('cols' => array(array('label' => 'Not completed', 'type' => 'string'),
                              array('label' => 'Completed', 'type' => 'string')),
              'rows' => array());

    while($row = mysql_fetch_row($result)) {
   $data['rows'][] = array('c' => array(array('v' => $row[0]), array('v' => $row[1])));
}    

echo json_encode($data);



  mysql_close($con);
?>

我确定我在对数组和查询做错了。此外，我正在考虑创建两个单独的SQL查询并将它们保存在两个php变量中; $ not_completed，$完成并在图表中填充它。不知道哪一个是我的问题的最佳选择。有人可以帮忙吗？

Answer 1

查看https://developers.google.com/chart/以查找Google图表的文档。如果你已经有变量$ complete和$ not_complete，你可以使用这个例子http://pastebin.com/qsB0hqmJ，你应该使用mysql_result改进你的mysql / php代码。

$tasks_completed = mysql_result(mysql_query("SELECT COUNT(*) FROM tasks..."),0);

使用Google Charts填充MySQL数据

1 个答案: