Javascript 谷歌图表API没有';你不能使用PHP吗?

Javascript 谷歌图表API没有';你不能使用PHP吗?,javascript,php,mysqli,google-api,Javascript,Php,Mysqli,Google Api,GoogleAPI图表没有接受我使用php所反映的值,有人能告诉我我做错了什么吗 这是用户选择要查看的职务图表的位置: <form class="form-horizontal form-select-chart" method="post" action="analytics.php"> <select name="chartname"> <?php $sel_

GoogleAPI图表没有接受我使用php所反映的值,有人能告诉我我做错了什么吗

这是用户选择要查看的职务图表的位置:

<form class="form-horizontal form-select-chart" method="post" action="analytics.php">
                <select name="chartname">
                <?php
                    $sel_user = "select * from tbl_jobs";
                    $run_user = mysqli_query($con, $sel_user);
                    while($row = mysqli_fetch_assoc($run_user))
                      {
                          echo "<option value=\"" . $row['Job_ID'] . "\">" . $row['Job_Name'] . "</option>";
                      }
                ?>
                </select>
                <br>
                <input type="submit" name="make_chart" value="Create the Chart" class="btn btn-primary" />
                </form>


以下是应处理数据的位置:

<script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>
<script type="text/javascript">
<?php
    if(isset($_POST['make_chart'])){
        $jid = mysqli_real_escape_string($con,$_POST['chartname']);
        $sel_user = "select * from tbl_jobs WHERE `ID` = '$jid'";
                    $run_user = mysqli_query($con, $sel_user);
                    while($row = mysqli_fetch_assoc($run_user))
                      {
                        $jname = $row['Job_Name'];
                        $overall_vacant = $row['Vacant'];   
                        $overall_filled = $row['Filled'];   
                      }
                      settype($overall_vacant, "integer");
                      settype($overall_filled, "integer");
    }
?>
  // Load the Visualization API and the corechart package.
  google.charts.load('current', {'packages':['corechart']});

  // Set a callback to run when the Google Visualization API is loaded.
  google.charts.setOnLoadCallback(drawChart);

  // Callback that creates and populates a data table,
  // instantiates the pie chart, passes in the data and
  // draws it.
  function drawChart() {
    // Create the data table.
    var data = new google.visualization.DataTable();
    data.addColumn('string', 'Topping');
    data.addColumn('number', 'Slices');
    data.addRows([
      ['Vacant Positions', <?php  echo $overall_vacant; ?>],
      ['Filled Positions', <?php  echo $overall_filled; ?>],
    ]);

    // Set chart options
    var options = {'title':'<?php echo $jname; ?>',
                   'width':400,
                   'height':300};

    // Instantiate and draw our chart, passing in some options.
    var chart = new google.visualization.PieChart(document.getElementById('chart_div'));
    chart.draw(data, options);
  }
</script>

//加载可视化API和corechart包。
load('current',{'packages':['corechart']});
//将回调设置为在加载Google Visualization API时运行。
google.charts.setOnLoadCallback(drawChart);
//创建并填充数据表的回调,
//实例化饼图,传入数据并
//画它。
函数绘图图(){
//创建数据表。
var data=new google.visualization.DataTable();
data.addColumn('string','Topping');
data.addColumn('number','Slices');
data.addRows([
[‘空缺职位’,],
[‘填补职位’,],
]);
//设置图表选项
变量选项={'title':'',
“宽度”:400,
‘高度’:300};
//实例化并绘制图表,传入一些选项。
var chart=new google.visualization.PieChart(document.getElementById('chart_div');
图表绘制(数据、选项);
}
以下是要显示的数据的位置:

    <?php 
        if(isset($_POST['make_chart'])){
        echo"<div align=\"center\" id=\"chart_div\"></div>"; 
        }

        unset($_POST['make_chart']));
    ?>

经过进一步的调查,我发现我的MySQL查询有问题!我应该用“工作ID”而不是“ID”,这是两个不同的东西

我的桌子是这样的:

ID | Job_ID | Job_Name | Vacant | Filled
1      0      Overall    0        0

这又导致了一张空白图表!我真傻!感谢MECU的提醒

你能展示一下提交表单后脚本的样子吗?我猜,
$jname
不是你想象的那样。或者,
选项。title
不应该有引号。@MECU它只是空的,显示标题“总体”,即使在选择表单中有其他选项