Chart.js-使用mysql和php从数据库获取数据

一尘不染

Chart.js-使用mysql和php从数据库获取数据

mysql

我正在尝试将静态数据转换为使用数据库结果。我将使用 MySQL 和 PHP 。

示例代码：

var randomScalingFactor = function(){ return Math.round(Math.random()*100)};
var lineChartData = {
    labels : ["January","February","March","April","May","June","July"],
    datasets : [
        {
            label: "My First dataset",
            fillColor : "rgba(220,220,220,0.2)",
            strokeColor : "rgba(220,220,220,1)",
            pointColor : "rgba(220,220,220,1)",
            pointStrokeColor : "#fff",
            pointHighlightFill : "#fff",
            pointHighlightStroke : "rgba(220,220,220,1)",
            data : [randomScalingFactor(),randomScalingFactor(),randomScalingFactor(),randomScalingFactor(),randomScalingFactor(),randomScalingFactor(),randomScalingFactor()]
        },
        {
            label: "My Second dataset",
            fillColor : "rgba(151,187,205,0.2)",
            strokeColor : "rgba(151,187,205,1)",
            pointColor : "rgba(151,187,205,1)",
            pointStrokeColor : "#fff",
            pointHighlightFill : "#fff",
            pointHighlightStroke : "rgba(151,187,205,1)",
            data : [randomScalingFactor(),randomScalingFactor(),randomScalingFactor(),randomScalingFactor(),randomScalingFactor(),randomScalingFactor(),randomScalingFactor()]
        }
    ]
    }
window.onload = function(){
    var ctx = document.getElementById("canvas").getContext("2d");
    window.myLine = new Chart(ctx).Line(lineChartData, {
        responsive: true
    });
}

以下是我的php / msql：

$result = mysql_query("SELECT COUNT(*) FROM customer WHERE month='january'") or die(mysql_error());
$row1 = mysql_fetch_array( $result );

$result = mysql_query("SELECT COUNT(*) FROM customer WHERE month='february'") or die(mysql_error());
$row2 = mysql_fetch_array( $result );

$result = mysql_query("SELECT COUNT(*) FROM customer WHERE month='march'") or die(mysql_error());
$row3 = mysql_fetch_array( $result );

如何使用count()我的MySQL查询中的那些并将其实现到chartjs上的数据集？我也希望标签也可以从我的MySQL查询中生成。我应该在jQuery代码内部循环数据集吗？

这是我正在使用的插件：http :
//www.chartjs.org/docs/#line-chart-
introduction

阅读 706

2020-05-17

共1个答案

一尘不染

请将您的php代码放入另一个文件中，api.php并使用 JSON*
格式$.ajax获取这些数据。要将数据转换为JSON格式的数据，应使用php函数。 *json_encode()

我提供了示例示例，您可以在这里看到。

请参考以下代码示例：

api.php
```
$arrLabels = array("January","February","March","April","May","June","July");
```
$arrDatasets = array(‘label’ => “My First dataset”,’fillColor’ => “rgba(220,220,220,0.2)”, ‘strokeColor’ => “rgba(220,220,220,1)”, ‘pointColor’ => “rgba(220,220,220,1)”, ‘pointStrokeColor’ => “#fff”, ‘pointHighlightFill’ => “#fff”, ‘pointHighlightStroke’ => “rgba(220,220,220,1)”, ‘data’ => array(‘28’, ‘48’, ‘40’, ‘19’, ‘86’, ‘27’, ‘90’));

$arrReturn = array(array(‘labels’ => $arrLabels, ‘datasets’ => $arrDatasets));

print (json_encode($arrReturn));
example.html
```
$.ajax({
```
type: ‘POST’,
url: ‘api.php’,
success: function (data) {
lineChartData = data;//alert(JSON.stringify(data));
var myLine = new Chart(document.getElementById(“canvas”).getContext(“2d”)).Line(lineChartData);

var ctx = document.getElementById(“canvas”).getContext(“2d”);
window.myLine = new Chart(ctx).Line(lineChartData, {responsive: true});
}
});

请注意，您应该传递价值randomScalingFactor()的api.php。

请检查并告知我是否需要其他帮助。

2020-05-17