码:
<?php
include('config.php');
?>
<html>
<head>
<style>
html,
body,
#myChart {
height: 100%;
width: 100%;
}
</style>
</head>
<body>
<select id="jid">
<option>Select Job</option>
<option value="jid1">Java Developer</option>
<option value="jid2">Dot Net Developer</option>
<option value="jid3">PHP Developer</option>
</select>
<div id='myChart'></div>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script src="https://cdn.zingchart.com/zingchart.min.js"></script>
<script>
zingchart.MODULESDIR = "https://cdn.zingchart.com/modules/";
ZC.LICENSE = ["569d52cefae586f634c54f86dc99e6a9", "ee6b7db5b51705a13dc2339db3edaf6d"];
</script>
<script>
$(document).ready(function(){
$("#jid").change(function(){
jid = $(this).val();
$.ajax({
type:"POST",
data:{"jid":jid},
url:"success.php",
success:function(data){
var myData = [data];
var myConfig = {
"graphset": [{
"type": "bar",
"title": {
"text": "Resume Tracking System"
},
"scale-x": {
"labels": ["Uploaded", "Shortlist", "Interview", "Final", "Offer"]
},
"series": [{
"values": myData
}]
}]
};
zingchart.render({
id: 'myChart',
data: myConfig,
height: "100%",
width: "100%"
});
}
});
});
});
</script>
</body>
</html>
success.php
<?php
include('config.php');
$jid = $_POST['jid'];
$sql = mysqli_query($con,"select * from test where jid='".$jid."'");
while($row = mysqli_fetch_array($sql))
{
echo $row['val'];
}
?>
我正在尝试使用jquery ajax生成更改下拉值的动态图。现在,当我调用ajax成功值时,我在这里做什么,然后它不生成图形但是手动输入值而不是data
我输入20
然后图形生成成功。我不知道为什么我做错了?请帮我。
谢谢
zingcharts文档说明系列数据应采用以下格式:
{
"type": "bar", /* or "vbar" */
"series": [
{"values":[20,40,25,50,15,45,33,34]},
]
}
console.log(myData)并确保它匹配