【发布时间】:2019-08-14 11:47:54
【问题描述】:
我有从数据库返回的数据,我正在使用 chartjs 在饼图上显示它们。
我希望用户选择一个特定的日期范围,当她点击过滤器时,它应该根据用户选择的日期在饼图上显示数据,live demo
这是我的解决方案。
HTML
from <input type="text" id = "firstdatepicker" name = "firstdatepicker">
to <input type="text" id = "lastdatepicker" name = "lastdatepicker">
<input type="button" name="filter" id="filter" value="Filter" class="btn btn-info" />
JS
$(document).ready(function(){
$(function() {
$( "#firstdatepicker" ).datepicker();
$( "#lastdatepicker").datepicker();
});
$('#filter').click(function(){
var from_date =$('#firstdatepicker').datepicker({ dateFormat: 'yy-mm-dd' }).val();
var to_date =$('#lastdatepicker').datepicker({ dateFormat: 'yy-mm-dd' }).val();
if(from_date != '' && to_date != ''){
$.ajax({
url:"https://meed.audiencevideo.com/admin/chats/stats.php",
type:"POST",
data:{from_date:from_date, to_date:to_date},
success:function(data){
console.log(data);
var session= data[0].sessions;
var yes = data[0].num_yes;
var no =data[0].num_no;
var ctx = document.getElementById("myPieChart");
var myChart = new Chart(ctx, {
type: 'pie',
data: {
labels: ["sessions","yes", "no"],
datasets: [{
label: 'Genders',
data: [session,yes, no],
backgroundColor: [
'rgba(255, 99, 132, 0.2)',
'rgba(54, 162, 235, 0.2)',
'rgba(54, 162, 235, 1)'
],
borderWidth: 1
}]
},
});
},
error: function () {
console.log('Something went wrong');
}
});
}
else
{
alert("Please Select Date");
}
});
})
这是从数据库中获取数据的php
if (isset($_POST['from_date']) && isset($_POST['to_date'])) {
$firstDate= $_POST['from_date'];
$lastDate= $_POST['to_date'];
$firstDate_new = date('Y-m-d', strtotime($firstDate));
$lastDate_new = date('Y-m-d', strtotime($lastDate));
$query = sprintf("SELECT count(*) as num_rows, datetime, count(distinct sid) as sessions, sum( targetbuttonname = 'yes' ) as num_yes, sum( targetbuttonname = 'no' ) as num_no from events WHERE datetime BETWEEN '{$firstDate_new}' AND '{$lastDate_new}'");
var_dump($query);
$result = $mysqli->query($query);
$data = [];
if(mysqli_num_rows($result) > 0)
{
while($row = mysqli_fetch_array($result))
{
$data[] = $row;
}
}
echo json_encode($data);
exit;
}else{
//query to get data from the table
$query = sprintf("SELECT count(*) as num_rows, count(distinct sid) as sessions, sum( targetbuttonname = 'yes' ) as num_yes, sum( targetbuttonname = 'no' ) as num_no from events;");
}
$result = $mysqli->query($query);
//loop through the returned data
$data = array();
foreach ($result as $row) {
$data[] = $row;
}
//free memory associated with result
$result->close();
//close connection
$mysqli->close();
//now print the data
print json_encode($data);
当我选择日期范围并单击过滤器时,我会在这样的控制台上获得所需的数据。
"SELECT count(*) as num_rows, datetime, count(distinct sid) as sessions, sum( targetbuttonname = 'yes' ) as num_yes, sum( targetbuttonname = 'no' ) as num_no from events WHERE datetime BETWEEN '2019-08-13' AND '2019-08-14'"
[
{
"0":"12",
"num_rows":"12",
"1":"2019-08-14",
"datetime":"2019-08-14",
"2":"12",
"sessions":"12",
"3":"2",
"num_yes":"2",
"4":"4",
"num_no":"4"
}
]
我收到来自 ajax something is wrong 的错误。当我尝试从POST 更改为GET 时,显示数据但旧数据。
我的代码做错了什么??
【问题讨论】:
-
所以你让它与 POST 一起工作?您显示的代码似乎是 POST 版本?
-
ajax 错误是什么?在php方面都正确吗?使用 curl 或 postman 等测试过这篇文章?
-
@Jonnix 在控制台上使用
post方法向我显示基于用户选择的日期范围的数据,但当我将其更改为GET我是获取其他数据 -
如果您删除
var_dump($query);是否有效? -
@SamuelTeixeira 你可以通过在控制台上打开网络自己看到它,meed.audiencevideo.com/admin/dashboard.php 和
post方法我得到了我想要的但是问题这些数据没有显示在饼图上 ajax 返回错误,something went wrong
标签: javascript php jquery mysql ajax