当使用php和sql时Google图表不会显示

问题描述：

我使用Google图表创建一个简单的图表。我在执行数据时使用PHP和MYSQL。我在寻找答案，并尝试，但它并没有显示图表当使用php和sql时Google图表不会显示

这里是我的test.php的文件

<html> 
    <head> 
    <script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script> 
    <script type="text/javascript"> 
     google.charts.load('current', {packages: ['corechart']}); 
     google.charts.setOnLoadCallback(drawChart); 

     function drawChart() { 
     // Define the chart to be drawn. 
     var data = google.visualization.arrayToDataTable([ 
     ['Specialization', 'Facultyno'], 
     <?php 
     require_once('../mydb_connect.php'); 
     $query="SELECT distinct(s.specializationname) as 'specialization', count(fs.facultyid) as 'facultyno' from thesisdb.specialization s join facultyspecialization fs on s.specializationid=fs.specializationid group by s.specializationID;"; 

     $result = mysqli_query($dbc,$query); 
     while($row = mysqli_fetch_array($result)){ 

     echo "['".$row['specialization']."',".$row['facultyno']."],"; 
     } 
     ?> 

     ]); 
     var options = {'title':'Number of Faculty Specializations', 
         'width':1300, 
         'height':900}; 

     } 
     // Instantiate and draw the chart. 
     var chart = new google.visualization.ColumnChart(document.getElementById('no_of_specialization')); 
     chart.draw(data, options); 
    } 
    </script> 
    </head> 

    <body> 
    <div id="no_of_specialization"/> 
    </body> 
</html>

在寻找这个问题，我是测试如果数据可以被执行，它的工作

这里是PHP代码

<?php 
     require_once('../mydb_connect.php'); 
     $query="SELECT distinct(s.specializationname) as specialization, count(fs.facultyid) as facultyno from thesisdb.specialization s join facultyspecialization fs on s.specializationid=fs.specializationid group by s.specializationID;"; 

     $result = mysqli_query($dbc,$query); 
     while($row = mysqli_fetch_array($result)) 
          { 
           echo "{$row['specialization']}"; 
           echo " - "; 
           echo "{$row['facultyno']}<br>"; 
          } 
     ?>

这是应该在谷歌字符显示的数据TS

人工智能 - 1

Biocomputation - 3

计算机和网络安全 - 2

人机交互 - 3

信息管理与分析 - 1

移动和互联网计算 - 1

真实世界计算 - 1

软件原理 - 2

理论计算机科学 - 1

我需要你的帮助。谢谢

答

你的代码中的每一件事都可以。但你正在使用一个额外的治疗大括号“}”。在这段代码后删除额外的括号'}'。

var options = {'title':'Number of Faculty Specializations', 
         'width':1300, 
         'height':900};

如果你遇到问题，找到额外治疗大括号“}”。只需复制并粘贴以下代码。

<html> 
<head> 
    <script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script> 
    <script type="text/javascript"> 
     google.charts.load('current', {packages: ['corechart']}); 
     google.charts.setOnLoadCallback(drawChart); 

     function drawChart() { 
      // Define the chart to be drawn. 
      var data = google.visualization.arrayToDataTable([ 
       ['Specialization', 'Facultyno'], 
       <?php 
       require_once('../mydb_connect.php'); 
       $query="SELECT distinct(s.specializationname) as 'specialization', count(fs.facultyid) as 'facultyno' from thesisdb.specialization s join facultyspecialization fs on s.specializationid=fs.specializationid group by s.specializationID;"; 

       $result = mysqli_query($dbc,$query); 
       while($row = mysqli_fetch_array($result)){ 

        echo "['".$row['specialization']."',".$row['facultyno']."],"; 
       } 
       ?> 

      ]); 
      var options = {'title':'Number of Faculty Specializations', 
       'width':1300, 
       'height':900}; 

     // Instantiate and draw the chart. 
     var chart = new google.visualization.ColumnChart(document.getElementById('no_of_specialization')); 
     chart.draw(data, options); 
     } 
    </script> 
</head> 

<body> 
<div id="no_of_specialization"/> 
</body> 
</html>

当使用php和sql时Google图表不会显示

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