Question

生成图表时出错

2013-08-06

268

php

在下面的代码中，我想生成饼图。但我收到此错误：未定义变量：mysqli，在第 29 行对非对象上的成员函数 query() 调用。请任何人帮助我纠正问题以获得解决方案。

<!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>

Answer 1

您不应结束 <form> ，直到所有字段都已打印。

您有一个

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

请先输出选择，然后关闭表单。否则，您的选择字段的值将不会提交。