生成圖表時出錯

在下面的代碼中，我想要生成餅圖，並且我得到了這個錯誤未定義變量：mysqli，在第29行的非對象上調用成員函數query（）。請任何人幫助我糾正問題得到解決方案。生成圖表時出錯

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
<html xmlns="http://www.w3.org/1999/xhtml"> 
<head> 
    <title>Pie Chart Demo (LibChart)- http://codeofaninja.com/</title> 
    <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-15" /> 
</head> 
<body> 

<?php 
    //include the library 
    include "libchart/libchart/classes/libchart.php"; 

    //new pie chart instance 
    $chart = new PieChart(500, 300); 

    //data set instance 
    $dataSet = new XYDataSet(); 

    //actual data 
    //get data from the database 

    //include database connection 
    include 'db_connect.php'; 

    //query all records from the database 
    $query = "select * from programming_languages"; 

    //execute the query 
    $result = $mysqli->query($query); 

    //get number of rows returned 
    $num_results = $result->num_rows; 

    if($num_results > 0){ 

     while($row = $result->fetch_assoc()){ 
      extract($row); 
      $dataSet->addPoint(new Point("{$name} {$ratings})", $ratings)); 
     } 

     //finalize dataset 
     $chart->setDataSet($dataSet); 

     //set chart title 
     $chart->setTitle("Tiobe Top Programming Languages for June 2012"); 

     //render as an image and store under "generated" folder 
     $chart->render("generated/1.png"); 

     //pull the generated chart where it was stored 
     echo "<img alt='Pie chart' src='generated/1.png' style='border: 1px solid gray;'/>"; 

    }else{ 
     echo "No programming languages found in the database."; 
    } 
?> 

</body> 
</html>

來源

2013-08-06 Moses Vickson

您不應該結束<form>，直到打印完所有字段。

你有

</form> 
<select>....</select>

請輸出選擇第一的然後關閉該窗體。否則，您選擇的字段的值將不會被引用。

來源

2013-08-06 07:54:51 steven

生成圖表時出錯

回答

相關問題