谷歌图不接受PHP变量


Google graph not accepting PHP Variables

我正在使用Google图表创建折线图。然而,在JavaScript部分,我尝试将我的PHP变量传递到数组中,但我收到了一条错误消息,内容是:

没有提供足够的列来绘制请求的图表。

我的代码:

<!-- Google Charts -->
<div id="chart_line_graph" style="width: 100%; height: 400px; margin: 0 auto;"></div>
<script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>
<?
    $titleArray = array(
                        'Week'
                        );
        $count = 0;
        //Get all headers
        $query = 'SELECT `keyword` FROM `'.$tableName.'` ORDER BY `keyword`';
        $results = mysqli_query($conn, $query);
        while($row = mysqli_fetch_assoc($results)){
            if($count < 5) //Get first 5 keywords
                $titleArray[] = $row['keyword'];
            $count++;
        }
        $data = array();
        $data[] = json_encode($titleArray);
        foreach($columns as $column){
            $weekData = array();
            $weekData[] = "'".$column."'";
            $query = 'SELECT `'.$column.'`  FROM `'.$tableName.'` ORDER BY `keyword`';
            $results = mysqli_query($conn, $query);
            $count = 0;
            while($row = mysqli_fetch_assoc($results)){
                if($count < 5)
                    $weekData[] = (Integer)$row[$column];
                $count++;
            }
            $data[] = json_encode($weekData);
        }
?> 
<script>
  google.charts.load('current', {'packages':['corechart']});
  google.charts.setOnLoadCallback(drawChart);
  function drawChart() {
    var data = google.visualization.arrayToDataTable([
    <?php
        foreach($data as $weekData){
            echo $weekData.',';
        }
      ?>    
    ],false);
    var options = {
      title: '<?echo ucwords($_SESSION['clientName']);?> Keywords',
      curveType: 'function',
      legend: { position: 'right' }
    };
    var chart = new google.visualization.LineChart(document.getElementById('chart_line_graph'));
    chart.draw(data, options);
  }
    </script>

每次运行它时,我都会收到相同的错误消息。我试着输出结果,并将其直接复制粘贴到JavaScript中,然后创建了图形。

有什么想法为什么不起作用吗?

编辑:

数据:

["Week","aker aberdeen","alma tavern","ancillary items","baby change sign","baby change units"],
["10_12_2015",36,11,37,30,48],
["17_12_2015",36,10,35,27,43],
["24_12_2015",26,11,35,26,44],
["31_12_2015",29,11,32,23,42],

我想明白了。出于某种原因,mysqli无法正常工作。我把它从程序化的方式改为面向对象的方式,一切都奏效了。我不知道为什么会这样,我只知道它有效。