I';我在使用CanvasJs连接MySQL时遇到了问题,我的值来了,但我的图表没有显示出来

I';我在使用CanvasJs连接MySQL时遇到了问题,我的值来了,但我的图表没有显示出来,mysql,canvasjs,Mysql,Canvasjs,正确对齐代码后,您会发现和标题前缺少两个逗号 您必须将构建的json传递给数据点 $'dbhost' = 'localhost'; $dbname = 'chart'; $dbuser = 'root'; $dbpass = ''; try{ $dbcon = new PDO("mysql:host={$dbhost};dbname={$dbname}",$dbuser,$dbpass); $dbcon->setAttribute(PDO::ATTR_ERRMODE,

正确对齐代码后,您会发现和标题前缺少两个逗号

您必须将构建的json传递给数据点

$'dbhost' = 'localhost'; $dbname = 'chart'; $dbuser = 'root'; $dbpass = '';

try{ 
    $dbcon = new PDO("mysql:host={$dbhost};dbname={$dbname}",$dbuser,$dbpass); 
    $dbcon->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
} 
catch(PDOException $ex)
    { die($ex->getMessage());
    } 
$stmt=$dbcon->prepare("SELECT * FROM contribution "); 
$stmt->execute(); 
while ($row=$stmt->fetch(PDO::FETCH_ASSOC))
  { 
   extract($row); 
   $json[] = $student; 
   $json2[] = (int)$contribution; 
  } 
echo json_encode($json); echo json_encode($json2); ?>

window.onload = function () 
    { 
    var chart = new CanvasJS.Chart("chartContainer", 
      { 
      animationEnabled: true
      , exportEnabled: true
       theme: "light1", // "light1", "light2", "dark1", "dark2" 
       title:{ text: "PHP Column Chart from Database" }
       , data: [{ type: "column", //change type to bar, line, area, pie, etc 
         dataPoints: }]
      }
      ); 
      chart.render(); 
    }
animationEnabled:真
,exportEnabled:true
,主题:“light1”//“light1”、“light2”、“dark1”、“dark2”
,标题:{text:“数据库中的PHP柱形图”}
,数据:[{type:“列”,
数据点:
}]
但由于这只是一个例子,你错过了什么


您必须检查$json是否属于

,代码是否正确对齐。您会发现和标题前缺少两个逗号

您必须将构建的json传递给数据点

$'dbhost' = 'localhost'; $dbname = 'chart'; $dbuser = 'root'; $dbpass = '';

try{ 
    $dbcon = new PDO("mysql:host={$dbhost};dbname={$dbname}",$dbuser,$dbpass); 
    $dbcon->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
} 
catch(PDOException $ex)
    { die($ex->getMessage());
    } 
$stmt=$dbcon->prepare("SELECT * FROM contribution "); 
$stmt->execute(); 
while ($row=$stmt->fetch(PDO::FETCH_ASSOC))
  { 
   extract($row); 
   $json[] = $student; 
   $json2[] = (int)$contribution; 
  } 
echo json_encode($json); echo json_encode($json2); ?>

window.onload = function () 
    { 
    var chart = new CanvasJS.Chart("chartContainer", 
      { 
      animationEnabled: true
      , exportEnabled: true
       theme: "light1", // "light1", "light2", "dark1", "dark2" 
       title:{ text: "PHP Column Chart from Database" }
       , data: [{ type: "column", //change type to bar, line, area, pie, etc 
         dataPoints: }]
      }
      ); 
      chart.render(); 
    }
animationEnabled:真
,exportEnabled:true
,主题:“light1”//“light1”、“light2”、“dark1”、“dark2”
,标题:{text:“数据库中的PHP柱形图”}
,数据:[{type:“列”,
数据点:
}]
但由于这只是一个例子,你错过了什么


你必须检查你的$JSON是否属于

你真的应该格式化你的代码。需要更多的问题描述你真的应该格式化你的代码。需要更多的问题描述你能告诉我,你添加了我的代码更改吗,你看到firefox或chrome的web开发者工具有错误吗?你能告诉我,你添加了我的代码更改吗,yoyu是否在firefox或chrome的web开发者工具中发现了错误?