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Javascript 用PHP、mySQL浏览JS动态数据_Javascript_Php_Mysql - Fatal编程技术网

Javascript 用PHP、mySQL浏览JS动态数据

Javascript 用PHP、mySQL浏览JS动态数据,javascript,php,mysql,Javascript,Php,Mysql,您好,我有来自CanvasJs的图表障碍 我只是用简单的代码来得到带有参数target和actual的简单图表,我在数据点中发现了错误:我认为问题只是错误的语句 这是我的错误代码: 数据点:[ 下面是我们如何在CanvasJS中显示MySQL数据的方法,请这样尝试 在这里,创建一个PHP服务,该服务以JSON格式返回数据。HTML页面向服务器发出AJAX请求并获取数据。获取数据后,它呈现一个图表 返回JSON数据的PHP服务 <?php header('Content-Type: app

您好,我有来自CanvasJs的图表障碍

我只是用简单的代码来得到带有参数target和actual的简单图表,我在数据点中发现了错误:我认为问题只是错误的语句

这是我的错误代码:

数据点:[


下面是我们如何在
CanvasJS
中显示
MySQL
数据的方法,请这样尝试

在这里,创建一个PHP服务,该服务以JSON格式返回数据。HTML页面向服务器发出AJAX请求并获取数据。获取数据后,它呈现一个图表

返回JSON数据的PHP服务

<?php

header('Content-Type: application/json');

$con = mysqli_connect("127.0.0.1","user","password","canvasjssampledb");

// Check connection
if (mysqli_connect_errno($con))
{
    echo "Failed to connect to DataBase: " . mysqli_connect_error();
}else
{
    $data_points = array();

    $result = mysqli_query($con, "SELECT * FROM sales");

    while($row = mysqli_fetch_array($result))
    {        
        $point = array("label" => $row['product'] , "y" => $row['total_sales']);

        array_push($data_points, $point);        
    }

    echo json_encode($data_points, JSON_NUMERIC_CHECK);
}
mysqli_close($con);

?>

获取数据和呈现图表的HTML页面

!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml" >
<head>
    <title></title>
    <script src="jquery.js"></script>
    <script src="canvasjs.js"></script>

    <script type="text/javascript">
        $(document).ready(function () {

            $.getJSON("data.php", function (result) {

                var chart = new CanvasJS.Chart("chartContainer", {
                    data: [
                        {
                            dataPoints: result
                        }
                    ]
                });

                chart.render();
            });
        });
    </script>
</head>
<body>

    <div id="chartContainer" style="width: 800px; height: 380px;"></div>

</body>
</html>
!DOCTYPE html>
$(文档).ready(函数(){
$.getJSON(“data.php”,函数(结果){
var chart=new CanvasJS.chart(“chartContainer”{
数据:[
{
数据点:结果
}
]
});
chart.render();
});
});
注:IE8-在v1.3及更高版本中受支持


您好,请再次帮助我,我的问题可以用您的答案解决,但是,如果我想用idEmp选择我的销售,那么我如何编写代码呢?例如$result=mysqli_查询($con,“select*FROM sales where idEmp='$id'));在我的示例dyanamic中如何生成$id?我不明白你的意思。但我希望这就是你的意思?!!
$result=mysqli\u query($con,“SELECT*FROM sales WHERE idEMP=”.id;
并将
mysqli_fetch_array
更改为
fetch_assoc
。很抱歉再次打扰您,请查看我的屏幕截图,如果您的示例代码使用sales table,我有一个具有不同类别值的图表,如果我有另一个图表,然后在idEmp中有where值,那么我可以从HTML页面获取JSON数据并渲染图表?
<?php

header('Content-Type: application/json');

$con = mysqli_connect("127.0.0.1","user","password","canvasjssampledb");

// Check connection
if (mysqli_connect_errno($con))
{
    echo "Failed to connect to DataBase: " . mysqli_connect_error();
}else
{
    $data_points = array();

    $result = mysqli_query($con, "SELECT * FROM sales");

    while($row = mysqli_fetch_array($result))
    {        
        $point = array("label" => $row['product'] , "y" => $row['total_sales']);

        array_push($data_points, $point);        
    }

    echo json_encode($data_points, JSON_NUMERIC_CHECK);
}
mysqli_close($con);

?>
!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml" >
<head>
    <title></title>
    <script src="jquery.js"></script>
    <script src="canvasjs.js"></script>

    <script type="text/javascript">
        $(document).ready(function () {

            $.getJSON("data.php", function (result) {

                var chart = new CanvasJS.Chart("chartContainer", {
                    data: [
                        {
                            dataPoints: result
                        }
                    ]
                });

                chart.render();
            });
        });
    </script>
</head>
<body>

    <div id="chartContainer" style="width: 800px; height: 380px;"></div>

</body>
</html>