PHP错误中的图表

时间:2015-02-27 13:05:10

标签: php mysql graph

我正在尝试从数据库中获取数据并生成图表。我使用的是phpgraphlib,http://www.ebrueggeman.com/phpgraphlib/documentation/tutorial-mysql-and-phpgraphlib。我从https://github.com/libgd/libgd/releases下载了libgd 还有phpgraphlib。我有PHP 5.5和php_gd2.dll扩展我认为它在php.ini文件中默认启用。

我收到2个错误:

  

警告:require_once(phpgraphlib.php):无法打开流:没有这样的   第3行的C:\ XAMPP \ htdocs \ graph.php中的文件或目录

     

致命错误:require_once():无法打开所需的'phpgraphlib.php'   (include_path ='。; C:\ XAMPP \ php \ PEAR')在C:\ XAMPP \ htdocs \ graph.php上   第3行

我也不知道路径和目录出了什么问题。有帮助吗?在此先感谢!!!

    <?php

      require_once("phpgraphlib.php");
      $graph=new PHPGraphLib(400,300);

     //connection to MySQL database

    $link =mysql_connect('localhost','root','') or die('Could not connect :'. mysql_error());

    //select db
    mysql_selected_db('mynewdb') or die('Could not select database');

   $dataArray=array();

   //get data from database
    $sql="SELECT year1,year2 * FROM scores WHERE crit='interesting semester' AND sid='13'";
    $result = mysql_query($sql) or die('Query failed: ' . mysql_error());

   if ($result) {
         while ($row = mysql_fetch_assoc($result)) {
            $crit=$row["crit"];
            $year1=$row["year1"];
            $year2=$row["year2"];
           //add to data array
            $dataArray[$crit]=$count;
         }
   }

       //configure graph
         $graph->addData($dataArray);
         $graph->setTitle("Interesting Semester");
         $graph->setGradient("lime", "green");
         $graph->setBarOutlineColor("black");
         $graph->createGraph();

   ?>

1 个答案:

答案 0 :(得分:0)

请将您的文件phpgraphlib.php放在./或C:\ XAMPP \ php \ PEAR