Google API 图表不接受我使用 php 回显的值,任何人都可以启发我做错了什么吗?
这是用户选择他们要查看的作业图表的地方:
<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 $jname; ?>',
'width':400,
'height':300};
// Instantiate and draw our chart, passing in some options.
var chart = new google.visualization.PieChart(document.getElementById('chart_div'));
chart.draw(data, options);
}
</script>
以下是要显示数据的位置:
<?php
if(isset($_POST['make_chart'])){
echo"<div align="center" id="chart_div"></div>";
}
unset($_POST['make_chart']));
?>
经过进一步调查,我发现我的MySQL查询有问题!我应该用"Job_ID"而不是"ID",这是两个不同的东西!
我的表格是这样的:
ID | Job_ID | Job_Name | Vacant | Filled
1 0 Overall 0 0
这反过来又导致了一个空白的图表!我真笨!感谢MECU的提醒!