程序流程:
1.加载页面后,图表将显示所有分支的总销售额。
2.当您从下拉按钮中选择一个分支时,图表显示的数据就是该特定分支的总销售额。

我的问题是,当我单击特定分支时,数据根本不会显示。

请参阅屏幕截图和代码以获取更多信息。

屏幕截图:
Screenshot of the page when it loads
Screenshot of the page when I choose a branch

PHP数据代码:

<div class="report-header">
    <h5 class="report-title">Total Yearly Sales</h5>
        <div class="branch-report">
            <select class="form-control" id="t-yearly">
                <option value="">Branch</option>
                <?php
                    require_once "connect.php";

                    $sql = "SELECT id,branch FROM tblLocation";
                    $result = mysqli_query($conn,$sql);

                    while ($row = mysqli_fetch_array($result)) {
                     echo "<option value='".$row['id']."'>".$row['branch']."</option>";
                     }
                    ?>
                </select>
            </div>
        </div>

        <div id="t-yearly-sales" style="height: 80%;"></div>
            <?php
                include "connect.php";

                $sql = "SELECT year, SUM(sales) AS sales FROM tblSales GROUP BY year";
                $result = mysqli_query($conn, $sql);

                $chart = '';
                while ($row = mysqli_fetch_array($result)){
                    $chart .= "{year:'".$row["year"]."', sales:".$row["sales"]."},";
                }
            ?>
            <script>
                new Morris.Bar({
                  element: 't-yearly-sales',
                  data: [<?php echo $chart; ?>],
                  xkey: 'year',
                  ykeys: ['sales'],
                  labels: ['Total Sales'],
                  hideHover: 'auto'
                });
            </script>


AJAX代码:

//Total Yearly Sales
$("#t-yearly").change(function(){
    var branch = $(this).val();
    $.ajax ({
        url:"fetch_yearly_sales.php",
        method: "POST",
        data: {branch:branch},
        success: function(branch_data){
             new Morris.Bar({
                element: 't-yearly-sales',
                data: [branch_data],
                xkey: 'year',
                ykeys: ['sales'],
                labels: ['Total Sales'],
                hideHover: 'auto'
             });
             console.log(branch);
        }
    });
});


Fetch_Yearly_Sales.php代码:

   <?php
require "connect.php";


$data = mysqli_real_escape_string($conn,$_POST['branch']);


if($data == ""){
    $output = "";
    $sql = "SELECT branch_id, year, SUM(sales) AS sales FROM tblSales GROUP BY year";
    $result = mysqli_query($conn, $sql);

    while ($row = mysqli_fetch_array($result)) {
        $chart .= "{year:'".$row["year"]."', sales:".$row["sales"]."},";
    }
    ob_clean();
    echo $output;
}
else{
    $output = "";
    $sql = "SELECT branch_id, year, SUM(sales) AS sales FROM tblSales WHERE branch_id='".$data."' GROUP BY year";
    $result = mysqli_query($conn, $sql);

    while ($row = mysqli_fetch_array($result)) {
        $chart .= "{year:'".$row["year"]."', sales:".$row["sales"]."},";
    }
    ob_clean();
    echo $output;
}
?>

最佳答案

您可以在PHP中尝试

if($data == ""){
$sql = "SELECT branch_id, year, SUM(sales) AS sales FROM tblSales GROUP BY year";
 $result = mysqli_query($conn, $sql);
 while ($row = mysqli_fetch_assoc($result)) {
   $chart = '{year:'.$row["year"].', sales:'.$row["sales"].'},';
 }
 echo $chart;
   } else {
  $sql = "SELECT branch_id, year, SUM(sales) AS sales FROM tblSales WHERE branch_id='".$data."' GROUP BY year";
$result = mysqli_query($conn, $sql);

while ($row = mysqli_fetch_array($result)) {
    $chart = '{year:'.$row["year"].', sales:'.$row["sales"].'}';
}
echo $chart;
}

关于javascript - 使用AJAX PHP显示数据,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/48879126/

10-12 14:25
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