这里是我的get_categories.php PHP代码:错误试图生成MySQL数据库JSON数组中的PHP
<?PHP
require_once('connection.php');
$query="SELECT * FROM categories";
$result = mysqli_query($connection,$query);
$return_arr = array();
while ($row = mysqli_fetch_array($result, mysqli_fetch_assoc)
{
$row_array['category'] = $row['category'];
$row_array['icon'] = $row['icon'];
array_push($return_arr,$row_array);
}
echo json_encode($return_arr);
?>
和connection.php
<?php
$servername = "localhost"; //replace it with your database server name
$username = "root"; //replace it with your database username
$password = "password"; //replace it with your database password
$dbname = "db_client";
// Create connection
$connection = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$connection) {
die("Connection failed: " . mysqli_connect_error());
}
?>
当我运行get_categories.php生成json数组..此错误出现
解析错误:语法错误,意外';'在第7行的C:\ xampp \ htdocs \ get_categories.php上
有人可以纠正我在做什么错吗?谢谢。
你缺少一个右')'您'while'线 –