使用AJAX PHP显示数据
使用AJAX PHP显示数据
提问于 2018-02-20 14:25:47
程序流程: 1.页面加载后，图表将显示所有分店的总销售额。2.当您从下拉按钮中选择一个分支时，图表显示的数据是该特定分支的总销售额。
我的问题是，当我点击一个特定的分支时，数据根本不会显示出来。
有关详细信息，请参阅屏幕截图和代码。
屏幕截图：Screenshot of the page when it loads Screenshot of the page when I choose a branch
数据的PHP代码：
<div class="report-header">
<h5 class="report-title">Total Yearly Sales</h5>
<div class="branch-report">
<select class="form-control" id="t-yearly">
<option value="">Branch</option>
<?php
require_once "connect.php";
$sql = "SELECT id,branch FROM tblLocation";
$result = mysqli_query($conn,$sql);
while ($row = mysqli_fetch_array($result)) {
echo "<option value='".$row['id']."'>".$row['branch']."</option>";
}
?>
</select>
</div>
</div>
<div id="t-yearly-sales" style="height: 80%;"></div>
<?php
include "connect.php";
$sql = "SELECT year, SUM(sales) AS sales FROM tblSales GROUP BY year";
$result = mysqli_query($conn, $sql);
$chart = '';
while ($row = mysqli_fetch_array($result)){
$chart .= "{year:'".$row["year"]."', sales:".$row["sales"]."},";
}
?>
<script>
new Morris.Bar({
element: 't-yearly-sales',
data: [<?php echo $chart; ?>],
xkey: 'year',
ykeys: ['sales'],
labels: ['Total Sales'],
hideHover: 'auto'
});
</script>AJAX代码：
//Total Yearly Sales
$("#t-yearly").change(function(){
var branch = $(this).val();
$.ajax ({
url:"fetch_yearly_sales.php",
method: "POST",
data: {branch:branch},
success: function(branch_data){
new Morris.Bar({
element: 't-yearly-sales',
data: [branch_data],
xkey: 'year',
ykeys: ['sales'],
labels: ['Total Sales'],
hideHover: 'auto'
});
console.log(branch);
}
});
});Fetch_Yearly_Sales.php代码：
<?php
require "connect.php";
$data = mysqli_real_escape_string($conn,$_POST['branch']);
if($data == ""){
$output = "";
$sql = "SELECT branch_id, year, SUM(sales) AS sales FROM tblSales GROUP BY year";
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_array($result)) {
$chart .= "{year:'".$row["year"]."', sales:".$row["sales"]."},";
}
ob_clean();
echo $output;
}
else{
$output = "";
$sql = "SELECT branch_id, year, SUM(sales) AS sales FROM tblSales WHERE branch_id='".$data."' GROUP BY year";
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_array($result)) {
$chart .= "{year:'".$row["year"]."', sales:".$row["sales"]."},";
}
ob_clean();
echo $output;
}
?>回答 2
Stack Overflow用户
发布于 2018-02-20 14:32:55
最后一个逗号无效。你真的应该考虑使用json_encode。
while ($row = mysqli_fetch_assoc($result)) {
$chart .= "{year:'".$row["year"]."', sales:".$row["sales"]."},";
}
echo $output;此外，您也没有打印任何内容：
$output = "";
echo $output;但是您可以使用$chart .= "";来预置数据。
json_encode示例：
$arr = array();
while ($row = mysqli_fetch_assoc($result)) {
$arr[] = $row;
}
echo json_encode( $arr );整个ajax示例：
require "connect.php";
$data = mysqli_real_escape_string($conn,$_POST['branch']);
if($data == ""){
$sql = "SELECT year, SUM(sales) AS sales FROM tblSales GROUP BY year";
$result = mysqli_query($conn, $sql);
$arr = array();
while ($row = mysqli_fetch_assoc($result)) {
$arr[] = $row;
}
echo json_encode( $arr );
}
else{
$sql = "SELECT year, SUM(sales) AS sales FROM tblSales WHERE branch_id='".$data."' GROUP BY year";
$result = mysqli_query($conn, $sql);
$arr = array();
while ($row = mysqli_fetch_assoc($result)) {
$arr[] = $row;
}
echo json_encode( $arr );
}Stack Overflow用户
发布于 2018-02-20 14:44:13
您可以在PHP中尝试此操作
if($data == ""){
$sql = "SELECT branch_id, year, SUM(sales) AS sales FROM tblSales GROUP BY year";
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_assoc($result)) {
$chart = '{year:'.$row["year"].', sales:'.$row["sales"].'},';
}
echo $chart;
} else {
$sql = "SELECT branch_id, year, SUM(sales) AS sales FROM tblSales WHERE branch_id='".$data."' GROUP BY year";
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_array($result)) {
$chart = '{year:'.$row["year"].', sales:'.$row["sales"].'}';
}
echo $chart;
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/48879126
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