2017-02-14 87 views
2

我是TeeChart for PHP的新手。在另一頁面渲染TeeChart for PHP

我發現的所有例子都在同一個php文件中創建它的圖表。

我想使用PHP腳本構建圖表,該腳本通過AJAX接收一些參數,並在生成AJAX調用的頁面上呈現圖表。

這可能嗎?這方面的例子?

此致敬禮。

Jayme Jeffman

回答

0

這裏一個簡單的例子:

getchart.php:

<?php 
    // get the q parameter from URL 
$q = $_REQUEST["q"]; 


//Includes 
include "../../sources/TChart.php"; 

$chart1 = new TChart(600,450); 
$chart1->getChart()->getHeader()->setText($q); 
$chart1->getAspect()->setView3D(false); 

$line1 = new Line($chart1->getChart()); 
$line1->setColor(Color::RED()); 
$chart1->addSeries($line1); 

// Speed optimization 
$chart1->getChart()->setAutoRepaint(false); 

for($t = 0; $t <= 10; ++$t) { 
    $line1->addXY($t, (10 + $t), Color::RED()); 
} 

$chart1->getChart()->setAutoRepaint(true); 

$chart1->render("chart1.png");  
$rand=rand(); 
echo "chart1.png?rand=".$rand; 
?> 

的test.html:

<!DOCTYPE html> 
<html> 
<head> 
<script> 
function showChart(str) { 
    var xmlhttp = new XMLHttpRequest(); 
    xmlhttp.onreadystatechange = function() { 
     if (this.readyState == 4 && this.status == 200) {   
      var image = document.getElementById("get_img"); 
      image.src = xmlhttp.responseText; 
     } 
    } 
    xmlhttp.open("GET", "getchart.php?q="+str, true); 
    xmlhttp.send(); 
} 
</script> 
</head> 
<body> 

<p><b>Enter the chart title below:</b></p> 
<form> 
Chart title: <input type="text" onkeyup="showChart(this.value)"> 
</form> 
<p><img id="get_img" /></p> 
</body> 
</html>