Question

我有PHP脚本，该脚本可连接到mySQL数据库，该数据库进行查询并返回数据，然后将其循环遍历并添加到数组中。这是我的php脚本：

<?php
// Database credentials
$dbHost = 'localhost';
$dbUsername = 'demo';
$dbPassword = 'demo';
$dbName = 'syndicationdashboard';
header('Access-Control-Allow-Origin: *');
header('Content-Type: application/json');

// Create connection and select db
$db = new mysqli($dbHost, $dbUsername, $dbPassword, $dbName);

// Get data from database
$sql = "select Priority,Automated,isAutomatable,isNotAutomatable from automation_progress where platform = 'Cox' and update_date in (select MAX(update_date) FROM automation_progress) order by priority";

$data = array();

if ($result = mysqli_query( $db, $sql )){
    while ($row = mysqli_fetch_assoc($result)) {
        $data[] = $row;
      }
    }

   $newData = array();
   $firstLine = true;

   foreach ($data as $dataRow)
   {
        if($firstLine)
        {
           $newData[] =array_keys($dataRow);
           $firstLine = false;
        }
      $newData[] = array_values($dataRow);
   }

#return $newData;
echo json_encode($newData);
mysqli_close($db);
?>

以下是此脚本的输出：

[["Priority","Automated","isAutomatable","isNotAutomatable"],["All","216","860","44"],["P1","213","567","34"],["P2","1","148","6"],["P3","2","136","3"],["P4","0","7","1"],["P5","0","2","0"]]

最后，我正在进行ajax调用，以从脚本中获取数据并将数据加载到Google图表中。

  <script type="text/javascript">
     google.charts.load('current', {'packages':['bar']});
     google.charts.setOnLoadCallback(drawChart);

     function drawChart() {

       jQuery.ajax({
           method: "POST",
           dataType: "JSON",
           url: "http://10.21.124.252:8080/automationBarDataCox.php",
           success: function (jsonData) {
             var data = google.visualization.arrayToDataTable(jsonData);

             var options = {
               chart: {
                 title: 'Automation Performance',
               },
               vAxes: {
                0: {title: 'Automated'},
              },
               legend: { position: 'left'},
               bars: 'vertical',
             };

             var chart = new google.charts.Bar(document.getElementById('cox_barchart'));

             chart.draw(data, google.charts.Bar.convertOptions(options));
           }
       });

     }
  </script>

加载图表时，由于数字返回为字符串而不是数字，因此返回的数据导致出现问题。有关如何解决此问题的任何想法？

Answer 1

mysqli_fetch_assoc()返回字符串的关联数组。如果要将值键入为整数，则必须做一些额外的工作：

if ($result = mysqli_query($db, $sql)) {
    while ($row = mysqli_fetch_assoc($result)) {
        if (count($data) === 0) {
            $data[] = array_keys($row);
        }

        $data[] = [
            $row['Priority'],
            intval($row['Automated']),
            intval($row['isAutomatable']),
            intval($row['isNotAutomatable']),
        ];
    }

    mysqli_close($db);

    echo json_encode($data);
}

这也消除了不必要的循环。

PHP脚本始终将数字值作为字符串返回

1 个答案: