Question

我正在使用Google Charts创建折线图。但是在JavaScript部分我尝试将我的PHP变量传递到数组中，但是我收到一条错误消息：

没有足够的列来绘制所请求的图表。

我的代码：

> final
  A B     df1.x1     df1.x2     df2.x1     df2.x2     df3.x1     df3.x2
1 1 Q 0.45840140 0.67613686 0.45840140 0.67613686 0.45840140 0.67613686
2 1 Q 0.04184920 0.49420098 0.04184920 0.49420098 0.04184920 0.49420098
3 2 Q 0.79841159 0.53494071 0.79841159 0.53494071 0.79841159 0.53494071
4 2 P 0.89847872 0.22059712 0.89847872 0.22059712 0.89847872 0.22059712
5 3 P 0.06430706 0.48076174 0.06430706 0.48076174 0.06430706 0.48076174
6 3 P 0.17436400 0.02307715 0.17436400 0.02307715 0.17436400 0.02307715

每次运行它都会收到相同的错误消息。我尝试输出结果并将其直接复制并粘贴到JavaScript中，然后创建了图表。

为什么这不起作用的任何想法？

编辑：

数据：

<!-- 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>

Answer 1

我明白了。出于某种原因，mysqli无法正常工作。我将它从程序方式改为面向对象方式，这一切都奏效了。我不知道为什么会这样，我才知道它有效。

Google图表不接受PHP变量

1 个答案: