Question

我正在尝试在Google图表中显示获取的数据，但是javascript中的PHP变量没有回显..如果我在javascript之外回显变量一切正常，所以我怀疑我的javascript中的某些内容一定是错误的..我仔细检查了一切，但我无法确定我们的问题在哪里..

代码没有显示任何错误！

这是浏览器源中的输出..

['Active Posts',$post_counts],['Categories',$category_counts],['Users',$user_counts],['Comments',$comment_counts],

有什么建议吗？

谢谢

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

        function drawChart() {
        var data = google.visualization.arrayToDataTable([
        ['Data', 'Count'],

            <?php 

            $data_text = ['Active Posts', 'Categories', 'Users', 'Comments'];
            $data_count = ['$post_counts', '$category_counts', '$user_counts', '$comment_counts'];

            for($i = 0; $i < 4; $i++) {

            echo "['$data_text[$i]'" . "," . "$data_count[$i]],";
            }

            ?>
          ]);

          var options = {
              chart: {
              title: '',
              subtitle: '',

              }
            };

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

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

       </script>

Answer 1

由于您已经使用PHP为JS输出数据，因此您只需使用json_encode将ARRAY转换为JSON数据。

<?php
$data = [
    'Data'          => 'Count',
    'Active Posts'  => $post_counts,
    'Categories'    => $category_counts, 
    'Users'         => $user_counts, 
    'Comments'      => $comment_counts  
];

$json = json_encode($data);
?>


var data = google.visualization.arrayToDataTable("<?php echo $json; ?>");

Answer 2

从php变量中删除引号，

使用此，

$data_count = [$post_counts, $category_counts, $user_counts, $comment_counts];

代替，

$data_count = ['$post_counts', '$category_counts', '$user_counts', '$comment_counts'];

Javascript中的PHP变量不起作用

2 个答案: