Javascript 谷歌图表API没有';你不能使用PHP吗?
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: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_
<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