最受欢迎

在尝试实现MVC模式时，无法获取要显示的google图形

Unable to get google graphs to display when trying to implement MVC pattern

本文关键字：获取 显示 图形 google 实现 MVC 模式 | 更新日期: 2023-09-27

希望有人能找到我的问题的解决方案。我正在尝试使用谷歌地图和PHP创建一个仪表板。在我尝试将其放入MVC模式之前，我能够显示图形，现在我得到的只是错误。（代码分为各自的文件夹Model、View和Controller，根目录中有一个索引文件）。这是我迄今为止的代码

索引

<?php
include_once ("Design/includes.php");
$controller = new Controller();
$controller->invoke();
?>

控制器

<?php
class Controller {
public $model;
public function __construct()
{
    $this->model = new tableModel();
}
public function invoke()
{
    $this->model -> tableData();
    include 'view/homeScreen.php';
}
}
?>

型号（这是针对表格的，另一个型号在同一文件夹中，但完全相同）

<?php
class tableModel {
public function tableData(){
    global $con;
$result = mysqli_query ( $con, "SELECT fk_EXECUTION_LAB_MACHINE as LAB_MACHINE,    COUNT(fk_EXECUTION_LAB_MACHINE) as TOTAL
                            FROM EXECUTION_QUEUE
                            WHERE fk_EXECUTION_LAB_MACHINE IS NOT NULL and WEEK (TEST_START_TIME) = WEEK( current_date ) -1 AND YEAR( TEST_START_TIME) = YEAR( current_date )
                            GROUP BY fk_EXECUTION_LAB_MACHINE order by count(fk_EXECUTION_LAB_MACHINE) desc;");

$rows = array();
$table = array();
$table['cols'] = array(
    // Labels for chart/column titles
    array('label' => 'LAB_MACHINE', 'type' => 'string'),
    array('label' => 'TOTAL', 'type' => 'number')
);
//Getting results array and populating table
$rows = array();
while($r = mysqli_fetch_array($result)) {
    $temp = array();
    $temp[] = array('v' => (string) $r['LAB_MACHINE']);
    $temp[] = array('v' => (int) $r['TOTAL']);
    $rows[] = array('c' => $temp);
}
//Get table in JSON format
$table['rows'] = $rows;
$jsonTable = json_encode($table);
echo $jsonTable;
return $jsonTable;
mysqli_close ( $con );

    }
}
?>

这是的视图

<html>
<head>
<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script type="text/javascript"
    src="http://underscorejs.org/underscore-min.js"></script>
<script type="text/javascript"
    src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
<script type="text/javascript">

var Dash = {
        widget : {},
        render : {
            pie : function() {
                var view = new google.visualization.DataView(Dash.data.pieEg);
                var options = {
                        animation : {duration : 300},
                        chartArea : {width : '100%', height : '70%'},
                        is3D: 'true',
                        title : 'Pie Chart',
                        tooltip : {showColorCode : true}
                        };
                        Dash.widget.pie.draw(view, options);
                },

            table : function() {
                Dash.widget.table = new google.visualization.ChartWrapper({
                    chartType: 'Table',
                    dataTable: Dash.data.tableEg,
                    options: {
                        allowHtml : true
                    },
                    containerId: 'table_div'
                });
                Dash.widget.table.draw();
            }
        },
        util : {
            bootstrap : function(){
                Dash.util.prepareData();
                Dash.render.table();
                Dash.util.createWidgets();
                Dash.util.refreshCharts();
            },
            createWidgets : function(){
                Dash.widget.pie = new google.visualization.PieChart(document.getElementById('pie_div'));
            },
            refreshCharts : function(tableEg){
                if (!tableEg || tableEg.length === 0) {
                    tableEg = [1,2];
                }
                Dash.render.pie(tableEg.slice(0));
            },
            prepareData : function(){
                Dash.data = {};
                var tableData = $.ajax({
                    url: "tableModel.php",
                    dataType:"json",
                    async: false
                }).responseText;
                Dash.data.tableEg = new google.visualization.DataTable(tableData);

                var chartData = $.ajax({
                    url: "pieModel.php",
                    dataType:"json",
                    async: false
                }).responseText;                       
                Dash.data.pieEg = new google.visualization.DataTable(chartData);
            }
        }
        };
        google.load('visualization', '1', {
            'packages':['corechart'],
            'callback': Dash.util.bootstrap
        });
        </script>
</head>
<body>
    <div id="row1">
        <div id="pie_div"
            style="width: 410px; height: 410px; display: inline-block"></div>
    </div>
    <div id="row2" style="width: 900px">
        <div id="table_div" style="width: 700px; display: inline-block"></div>
        <div id="gauge" style="display: inline-block; float: right"></div>
    </div>
</body>
</html>

正如我之前所说，当我把视图和表数据放在一个文件夹中时，这些图工作得很好，但现在我把它放进了MVC风格，它不再工作了。这是我得到的错误

{"cols":[{"label":"LAB_MACHINE","type":"string"},{"label":"TOTAL","type":"number"}],"rows":[{"c":[{"v":"LIT-QAVIST-20"},{"v":1}]},{"c":[{"v":"LIT-QAVIST-25"},{"v":1}]}]}Warning: mysqli_query(): Couldn't fetch mysqli in C:'Users'n0237569'workspace'Dashboard MVC'Design'model'tableModel.php on line 11

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in C:'Users'n0237569'workspace'Dashboard MVC'Design'model'tableModel.php on line 28 {"cols":[{"label":"LAB_MACHINE","type":"string"},{"label":"TOTAL","type":"number"}],"rows":[]}

Warning: include(view/homeScreen.php): failed to open stream: No such file or directory in C:'Users'n0237569'workspace'Dashboard MVC'Design'controller'Controller.php on line 17

Warning: include(): Failed opening 'view/homeScreen.php' for inclusion (include_path='.;C:'php'pear') in C:'Users'n0237569'workspace'Dashboard MVC'Design'controller'Controller.php on line 17

正如您从第一个错误中看到的那样，它以json形式返回数据，但我不确定接下来会发生什么。

这是包含文件夹的内容，因为最后两个错误引用了

包括

<?php
require_once "Design/config.php";
require_once "Design/connection.php";
require_once "Design/model/pieModel.php";
require_once "Design/model/tableModel.php";
require_once "Design/controller/Controller.php";
?>

如有任何帮助，将不胜感激

感谢

Warning: include(view/homeScreen.php): failed to open stream: No such file or directory in C:'Users'n0237569'workspace'Dashboard MVC'Design'controller'Controller.php on line 17

Warning: include(): Failed opening 'view/homeScreen.php' for inclusion     (include_path='.;C:'php'pear') in C:'Users'n0237569'workspace'Dashboard     MVC'Design'controller'Controller.php on line 17

这些错误是由您的控制器中的这条线引起的

include 'view/homeScreen.php';

因为在你的控制器中，你在Controller文件夹中，所以你需要查找一个级别，即

include '../view/homeScreen.php';

在您的表格中型号

return $jsonTable;
mysqli_close ( $con );

tableModel底部的mysqli_close( $con );部分是无用的，因为你已经在它上面的return了。当请求完成时，PHP会自动关闭连接，但如果不再需要，可能值得自己关闭它。

您应该在MySQL函数周围封装一些逻辑来调试查询。

你可以试试

if( !$result ) { die(mysqli_error($con));

您应该考虑使用PDO，并查看一些现有的MVC框架，以更好地了解MVC模式，以及它们如何处理路由和自动加载类等。

相关文章：
最新更新
www.56WenDa.com 2012-2023 | php问答网 | php学习