错误试图生成MySQL数据库JSON数组中的PHP

Question

这里是我的get_categories.php PHP代码：

<?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。

Answer 1

while ($row = mysqli_fetch_array($result, mysqli_fetch_assoc)
应该是：
while ($row = mysqli_fetch_array($result, mysqli_fetch_assoc))

你缺少一个括号