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我想调用一个函数作为图像 src。我正在使用 phpgraphlib 绘制图形。我试过这样但得到垃圾值。`

        <?php
         include("phpgraphlib.php");
         function  kk()
         { 
        $graph=new PHPGraphLib(1000,1000); 
        include("db_connect.php"); 
        $dataArray=array();
        $graph_array=array();
        $sql="SELECT name,mark,entered_time FROM student ";
        $result = mysql_query($sql,$con) ;
        if ($result) {
      while ($row = mysql_fetch_assoc($result)) {

    $without_comma_value=explode(',', $row['mark']);
    $count=count($without_comma_value);

   for($i=0;$i<$count;$i++)
   {
   $Val_onebyone= $without_comma_value[$i];
   $num=$i+1;
   $dataArray[$num]=$Val_onebyone; 
         } 

       }
        }

      $graph->setBackgroundColor("#F78181");
      $graph->addData($graph_array);
      $graph->setBars(false);
      $graph->setLine(true);
      $graph->setupYAxis(20, 'black');
      $graph->setupXAxis(20, 'black');
      $graph->setTextColor('black');
      $graph->setDataPoints(true);
      $graph->setDataPointColor('maroon');
      $graph->setLineColor('maroon');
      $graph->createGraph();}
       ?>
  <html>
  <form>
      <div align="center">
          <table><tr>
              <td valign="mid"><b>SpO2</b></td>
              <td align="center">
                  <img src="<?php echo kk(); ?>" />
              </td>
          </tr></table>
      </div>
 <form>

`当在另一个 php 页面上编写此图形创建函数并尝试将该页面称为 img src 时，图形正在正常运行。但我想将其称为 function 。请帮忙

4

1 回答 1

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createGraph()直接写入 PHP 的输出管道。您需要缓存它并回显它。

更改最后两行

  $graph->setLineColor('maroon');
  $graph->createGraph();}

进入这个：

  $graph->setLineColor('maroon');
  ob_start();
  $graph->createGraph();
  $out = ob_get_clean();
  echo $out;
  }
于 2014-10-07T11:02:16.183 回答