我正在使用 datefrom 和 dateto 制作一个下拉菜单，然后显示这些日期之间的日期，但它没有显示，并且我不断在控制台中收到此错误消息。
VM652:1
Uncaught SyntaxError: Unexpected token '<', "<br />
<b>"... is not valid JSON
at JSON.parse (<anonymous>)
at Object.success (<anonymous>:18:30)
at c (jquery-3.6.0.min.js:2:28327)
at Object.fireWith [as resolveWith] (jquery-3.6.0.min.js:2:29072)
at l (jquery-3.6.0.min.js:2:79901)
at XMLHttpRequest.<anonymous> (jquery-3.6.0.min.js:2:82355)
这是脚本：
<script>
// Add an event liste ner for the form submission
document.querySelector('#submit-button').addEventLi stener('click', function() {
// Prevent the default form submission behavior
event.preventDefault();
// Get the selected values of the "fromdate" and "todate" dropdown menus
var fromDate = $('#fromdate').val();
var toDate = $('#todate').val();
// Send the AJAX request to the PHP script
$.ajax({
type: "POST",
url: "functions/search-sales_function.php",
data: { fromDate: fromDate, toDate: toDate },
success: function (response) {
// Parse the JSON response
var salesData = JSON.parse(response);
// Loop over the sales data and insert it into the table
for (var i = 0; i < salesData.length; i++) {
var date = salesData[i]['date'];
var sales = salesData[i]['total_sales'];
$('#sales-table tbody').append(`
<tr>
<td>${date}</td>
<td>${sales}</td>
</tr>`
);
}
},
});
});
</script>
这是 php 文件：
<?php
include "db_conn.php";
// Get the values of the "fromdate" and "todate" dropdown menus
$fromDate = $_POST['fromdate'];
$toDate = $_POST['todate'];
// Check the connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
// Write a SELECT statement to get the sales data between the selected dates
$sql = "SELECT date, SUM(sales) as total_sales FROM sales WHERE date BETWEEN '$fromDate' AND '$toDate' GROUP BY date ORDER BY date ASC";
// Execute the SELECT statement and fetch the results
$result = mysqli_query($conn, $sql);
// Loop over the results and add a row to the table for each date
while ($row = mysqli_fetch_assoc($result)) {
$date = $row['date'];
$sales = $row['total_sales'];
echo "<tr>
<td>$date</td>
<td>$sales</td>
</tr>";
}
// Close the connection
mysqli_close($conn);
?>
我想显示日期在所选日期之间的行，但控制台日志中不断显示错误，并且表格未显示
该错误与您的ajax方法的响应错误有关。该代码发送了
ajax请求的结果。您应该在 php 文件中使用echo json_encode($data);。$data = []; while ($row = mysqli_fetch_assoc($result)) { $data[] = $row; echo json_encode($data); }